Binomial Theorem

$$64^{64} = (63 + 1)^{64} = 63\lambda_1 + 1$$

$${\left({\left(64\right)}^{(64)}\right)}^{(64)}=(63\lambda +1{)}^{64}=63{\lambda }_2+1$$

$$\begin{gathered} T_5, T_6 \& T_7 are in A.P. \\ 2T_6=T_5+T_7 \end{gathered}$$

\(2 \times{ }^{n+4} C_5={ }^{n+4} C_4+{}^{n+4}C_{6}\)

\(\Rightarrow \quad 2 \times \frac{(n+4)!}{5! \times (n-1)!}=\frac{(n+4)!}{4! \times n!}+\frac{(n+4)!}{6! \times (n-2)!}\)

\(\begin{aligned}
& \Rightarrow \frac{2 }{5(n-1)}=\frac{1}{n(n-1)}+\frac{1}{6 \times 5} \\
& \Rightarrow \frac{2}{5(n-1)}=\frac{1}{n(n-1)}+\frac{1}{30} \\
& \Rightarrow 2 n=5+\frac{n(n-1)}{6}
\end{aligned}\)

\(\begin{aligned}
&\begin{array}{r}
\quad 12 n=30+n^2-n \\
\Rightarrow \quad n^2-13 n+30=0 \\
\Rightarrow \quad(n-10)(n-3)=0 \\
\quad n=10,(3) \\
\end{array}\\
&=\text { greatest binomial coeff }\\
&={}^7 C_3=\frac{7!}{3!.4!}=7 \times 5=35
\end{aligned}\)

Let \(S=26\left(\frac{2^3}{3}({}^{12}C_2)+\frac{2^5}{5}({}^{12}C_4)+\frac{2^7}{7}({}^{12}C_6)+\cdots+\frac{2^{13}}{13}({}^{12}C_{12})\right)\)

\(S=26\sum_{r=1}^{6}\frac{2^{2r+1}}{2r+1}({}^{12}C_{2r})\)

Using \(\frac{{}^{12}C_{2r}}{2r+1}=\frac{{}^{13}C_{2r+1}}{13}\), we get:

\(S=26\sum_{r=1}^{6}\frac{2^{2r+1}}{13}({}^{13}C_{2r+1})\)

\(S=2\sum_{r=1}^{6}2^{2r+1}({}^{13}C_{2r+1})\)

\(S=2\left[\sum_{r=0}^{6}2^{2r+1}({}^{13}C_{2r+1})-2({}^{13}C_1)\right]\)

\(S=2\sum_{r=0}^{6}2^{2r+1}({}^{13}C_{2r+1})-52\)

\(\sum_{r=0}^{6}{}^{13}C_{2r+1}2^{2r+1}=\frac{(1+2)^{13}-(1-2)^{13}}{2}\)

\(S=2\cdot\frac{3^{13}-(-1)^{13}}{2}-52\)

\(S=3^{13}+1-52=3^{13}-51\)

Given \(S=3^{13}-\alpha\), so \(\alpha=51\)

General term, $T_{r+1}={}^{9}C_r{\left(x^{2/3}\right)}^{9-r }{\left(\frac{x^{-2/5}}{2}\right)}^r$

$={}^9\mathrm{C}_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{r}}{\left(\mathrm{r}\right)}^{\left(6-\frac{2\mathrm{r}}{3}-\frac{2\mathrm{r}}{5}\right)}$

for coefficient of $x^{2/3}$,

put $6-\frac{2r}{3}-\frac{2r}{5}=\frac{2}{3}$

\(\Rightarrow r = 5\)

Coefficient of $x^{2/3}$ $={}^{9}C_5{\left(\frac{1}{5}\right)}^5$

For coefficient of $x^{-2/5}$

put $6-\frac{2r}{3}-\frac{2r}{5}=-\frac{2}{3}$

\(\Rightarrow r = 6\)

Coefficient of $x^{-2/5}$ is ${}^{9}C_6{\left(\frac{1}{2}\right)}^6$

sum $={}^{9}C_5{\left(\frac{1}{2}\right)}^5+{}^{9}C_6{\left(\frac{1}{2}\right)}^6=\frac{21}{4}$

General term, $T_{r+1}={}^{9}C_r{\left(x^{2/3}\right)}^{9-r }{\left(\frac{x^{-2/5}}{2}\right)}^r$

$={}^9\mathrm{C}_{\mathrm{r}}{\left(\frac{1}{2}\right)}^{\mathrm{r}}{\left(\mathrm{r}\right)}^{\left(6-\frac{2\mathrm{r}}{3}-\frac{2\mathrm{r}}{5}\right)}$

for coefficient of $x^{2/3}$,

put $6-\frac{2r}{3}-\frac{2r}{5}=\frac{2}{3}$

\(\Rightarrow r = 5\)

Coefficient of $x^{2/3}$ $={}^{9}C_5{\left(\frac{1}{5}\right)}^5$

For coefficient of $x^{-2/5}$

put $6-\frac{2r}{3}-\frac{2r}{5}=-\frac{2}{3}$

\(\Rightarrow r = 6\)

Coefficient of $x^{-2/5}$ is ${}^{9}C_6{\left(\frac{1}{2}\right)}^6$

sum $={}^{9}C_5{\left(\frac{1}{2}\right)}^5+{}^{9}C_6{\left(\frac{1}{2}\right)}^6=\frac{21}{4}$

$P_n={\sum }_{r=0}^n\frac{{ }^n{C}_r{(-2)}^r}{r+1}$

$={\sum }_{r=0}^n\frac{1}{\left(n+1\right)}{ }^{n+1}{C}_{r+1}{(-2)}^r$

$=\frac{-1}{2\left(n+1\right)}{\sum }_{r=0}^n{ }^{n+1}{C}_{r+1}{(-2)}^{r+1}$

$=\frac{-1}{2\left(\text{n}+1\right)}\left[{(1-2)}^{\text{n}+1}-1\right]$

${\text{⇒P}}_{\text{n}}=\frac{1}{2\left(\text{n}+1\right)}\left[1-{(-1)}^{\text{n}+1}\right]$

$P_{2n}=\frac{1}{2\left(2n+1\right)}\left[1-{(-1)}^{2n+1}\right]$

$P_{2n}=\frac{1}{2n+1}$

\(\sum_{n=1}^{25} \frac{1}{P_{2 n}}=\sum_{n=1}^{25}(2 n+1)=3+5+\ldots . .+51=\frac{25}{2}[51+3]=25 \times 27=675\)

\(\begin{aligned}
&\begin{aligned}
& \text { coeff. of } \mathrm{x} \Rightarrow \mathrm{p}-\mathrm{q}=1 \Rightarrow \mathrm{p}=1+\mathrm{q} \\
& \text { coeff. of } \mathrm{x}^2 \Rightarrow{ }^q C_2-p q+{ }^p C_2=-2 \\
& \frac{p(p-1)}{2}-p q+\frac{(q-1) q}{2}=-2 \\
& \Rightarrow \mathrm{p}^2-\mathrm{p}-2 \mathrm{pq}+ \mathrm{q}^2-\mathrm{q}=-4 \\
& \Rightarrow \mathrm{q}=2 \\
& \mathrm{p}=3
\end{aligned}\\
&\text { So, } p^2+q^2=3^2+2^2=13
\end{aligned}\)

\(
\begin{aligned}
& (1-x)\left[(1-x)\left(1+x+x^2\right)\right]^{2007} \\
& =(1-x)\left(1-x^3\right)^{2007} \\
& =\left(1-x^3\right)^{2007}-x\left(1-x^3\right)^{2007}
\end{aligned}
\)

\([(1- \left.x^3\right)^{2007}\) contains \(3 \lambda\) types of exponents while \(x\left(1-x^3\right)^{2007}\) will have \((3 \lambda+1)\) type while 2012 is
\((3 \lambda+2)\) type] that is not possible \(\Rightarrow 0\)
Coefficient of \(x^{2012}\) in \(\left(1-x^3\right)^{2007}=0\)
Coefficient of \(x^{2011}\) in \(\left(1-x^3\right)^{2007}=0\)
\(\Rightarrow\) Coefficient of \(x^{2012}\) in \((1-x)^{2008}\left(1+x+x^2\right)^{2007}=0\)

Given expansion:

$\left(x^4+2x^2+1\right)\left({}^{n}C_0{x}^0+{}^{n}C_1{x}^1+{}^{n}C_2{x}^2+{}^{n}C_3{x}^3+\dots \right)$

Coefficient $x\Rightarrow {}^{n}C_1$,
coeff. of $x^2\Rightarrow 2+{}^{n}C_2$$=2+\frac{n(n-1)}{2}$

Coeff. of $x^3=2.{}^{n}C_1+{}^{n}C_3$

$=2n+\frac{n(n-1)(n-2)}{6}$

Now according to question

$$\begin{gathered} n+2n+\frac{n(n-1)(n-2)}{6}=2\left[2+\frac{n(n-1)}{2}\right] \\ \Rightarrow 3n+\frac{n(n-1)(n-2)}{6}=4+n(n-1) \\ \Rightarrow n^3-9n^2+26n-24=0 \\ \Rightarrow n=2,3,4 \end{gathered}$$

Now checking for \(n = 2\)

Coeff. of \(x=2\), Coeff. of \(x^2=3\), Coeff. of \(x^3=4\)

are in A.P.
$\Rightarrow n=2$ is also the correct choice
Required sum of values of \(n=2+3+4=9\)

Let $S={(1+x)}^{1000}+x{(1+x)}^{999}+x^2{(1+x)}^{998}+\dots .+x^{1000}$

$={(1+x)}^{1000}\frac{\left(1-{\left(\frac{x}{1+x}\right)}^{1001}\right)}{1-\frac{x}{1+x}}$

$={(1+x)}^{1001}-x^{1001}$

The sum of the coefficients of \(x^{499}\) and \(x^{500}\)

${=}^{1001}{C}_{499}{+}^{1001}{C}_{500}{=}^{1002}{C}_{500}$

$$\begin{gathered} S=\sum _{r=50}^{100}\frac{{}^{100}C_r}{r+1}=\sum _{r=50}^{100}\frac{1}{r+1}·\frac{r+1}{101}·{}^{101}C_{r+1} \\ S=\frac{1}{101}\sum _{r=50}^{100}{}^{101}C_{r+1} \\ \because \sum _{r=0}^n{}^{n}C_r=2^n , {}^{n}C_r={}^{n}C_{n-r} \\ \Rightarrow \sum _{r=0}^{50}{}^{101}C_r=\sum _{r=51}^{101}{}^{101}C_r=\frac{2^{101}}{2}=2^{100} \\ \therefore S=\frac{1}{101}\times \frac{2^{101}}{2}=\frac{2^{100}}{101} \end{gathered}$$

${(\sqrt[4]{3}+\sqrt[3]{4})}^{12}$

q = 27+256 = 283
Now, ${}^{12}\mathrm{C}_{\mathrm{r}-1}+{}^{12}\mathrm{C}_{\mathrm{r}+1}=2·{}^{12}\mathrm{C}_{\mathrm{r}}$
$\frac{12!}{(r-1)!(13-r)!}+\frac{12!}{(r+1)!(11-r)!}=2·\frac{12!}{r!(12-r)!}$
$\frac{1}{(13-r)(12-r)}+\frac{1}{(r+1)\left(r\right)}=\frac{2}{r(12-r)}$
$2r^2-24r+156=26r+26-2r^2-2r$
$2{\mathrm{r}}^2-24\mathrm{r}+65=0$
No integral value of r.
$$\begin{gathered} p = 0 \\ p+q = 283 \end{gathered}$$

Let $k = r + 1$. When $r = 0$, $k = 1$; when $r = 10$, $k = 11$. The sum becomes:

$$\sum_{k=1}^{11} \binom{11}{k}$$

Using the binomial theorem, $\sum_{k=0}^{11} \binom{11}{k} = 2^{11}$. Thus:

$$\sum_{k=1}^{11} \binom{11}{k} = 2^{11} - \binom{11}{0} = 2^{11} - 1$$

Multiplying by 10:

$$10 \sum_{r=0}^{10} \binom{11}{r+1} = 10(2^{11} - 1)$$

Again, let $k = r + 1$, so $r = k - 1$. The sum becomes:

$$\sum_{k=1}^{11} \frac{\binom{11}{k}}{10^{k-1}} = 10 \sum_{k=1}^{11} \frac{\binom{11}{k}}{10^k}$$

Using the binomial theorem for $\left(1 + \frac{1}{10}\right)^{11}$:

$$\left(1 + \frac{1}{10}\right)^{11} = \sum_{k=0}^{11} \binom{11}{k} \left(\frac{1}{10}\right)^k \implies \sum_{k=0}^{11} \frac{\binom{11}{k}}{10^k} = \left(\frac{11}{10}\right)^{11}$$

Subtracting the $k=0$ term:

$$\sum_{k=1}^{11} \frac{\binom{11}{k}}{10^k} = \left(\frac{11}{10}\right)^{11} - 1$$

Multiplying by 10:

$$10 \sum_{k=1}^{11} \frac{\binom{11}{k}}{10^k} = 10\left(\frac{11^{11}}{10^{11}} - 1\right) = \frac{11^{11}}{10^{10}} - 10$$

Substitute the results back into the original expression:

$$10(2^{11} - 1) - \left(\frac{11^{11}}{10^{10}} - 10\right) = 10 \cdot 2^{11} - 10 - \frac{11^{11}}{10^{10}} + 10 = 10 \cdot 2^{11} - \frac{11^{11}}{10^{10}}$$

Notice that $10 \cdot 2^{11} = \frac{(20)^{11}}{10^{10}}$ (since $20^{11} = (2 \cdot 10)^{11} = 2^{11} \cdot 10^{11}$, so dividing by $10^{10}$ gives $2^{11} \cdot 10$). Thus:

$$10 \cdot 2^{11} - \frac{11^{11}}{10^{10}} = \frac{20^{11} - 11^{11}}{10^{10}}$$

Comparing with the given form $\frac{\alpha^{11} - 11^{11}}{10^{10}}$, we see $\alpha = 20$.

The first term is $\frac{x+1}{x^{2/3} - x^{1/3} + 1}$.

Substituting this into the first term: $\frac{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$

The second term is $\frac{x-1}{x - x^{1/2}}$.

Substituting these: $\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}=\frac{\sqrt{x}+1}{\sqrt{x}}$$=\frac{x^{1\mathrm{/}2}+1}{x^{1\mathrm{/}2}}=1+x^{-1\mathrm{/}2}$

${\left(\right(x^{1\mathrm{/}3}+1)-(1+x^{-1\mathrm{/}2}\left)\right)}^{10}$$={(x^{1\mathrm{/}3}-x^{-1\mathrm{/}2})}^{10}$

$$\begin{gathered} T_{r+1}=\left(\genfrac{}{}{0px}{}{10}{r}\right)\left(x^{1\mathrm{/}3}{)}^{10-r}\right(-x^{-1\mathrm{/}2}{)}^r \\ =\left(\genfrac{}{}{0px}{}{10}{r}\right)(-1{)}^r{x}^{\frac{10-r}{3}-\frac{r}{2}} \end{gathered}$$

For the term to be independent of $x$, the exponent of $x$ must be zero:

$\frac{10-r}{3} - \frac{r}{2} = 0$ :

$2(10-r) - 3r = 0$

$20 - 2r - 3r = 0$

$20 = 5r \implies r = 4$

Substitute $r=4$ into the expression for

$T_{r+1}$: $T_5=\left(\genfrac{}{}{0px}{}{10}{4}\right)(-1{)}^4=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\times 1=10\times 3\times 7=210$

\(\left(2 x^2+\frac{1}{x}\right)^{10}, x \neq 0 \)
\( T_{r+1}={ }^{10} C_r\left(2 x^2\right)^{10-r}\left(\frac{1}{x}\right)^r \)
\( T_{r+1}={ }^{10} C_r(2)^{10-r} \cdot(x)^{20-3 r}\)
We need to find coefficient of \(x^2\)

\(\therefore 20-3 r=2 \)
\( 3 r=18 \Rightarrow r=6 \)
\( \therefore\) Coefficient of \( x^2={ }^{10} C_6(2)^{10-6} \)
\( ={ }^{10} C_6(2)^4=210 \times 16=3360\)

Middle term in the expansion of \((1+\alpha x)^{26}\):

\(\mathrm{T}_{14}={ }^{26} \mathrm{C}_{13} \alpha^{13} x^{13}\)

For \((1-\alpha x)^{28}\),

Middle term \(\mathrm{T}_{15}={ }^{28} \mathrm{C}_{14} \alpha^{14} x^{14}\)

Given that coefficients of middle terms are equal:

\(\Rightarrow{ }^{26} \mathrm{C}_{13} \alpha^{13}={ }^{28} \mathrm{C}_{14} \alpha^{14}\)

\(\Rightarrow{ }^{26} \mathrm{C}_{13}={ }^{28} \mathrm{C}_{14} \alpha\)

\(\Rightarrow \alpha=\frac{7}{27}\)

$P_n={\sum }_{r=0}^n\frac{{ }^n{C}_r{(-2)}^r}{r+1}$

$={\sum }_{r=0}^n\frac{1}{\left(n+1\right)}{ }^{n+1}{C}_{r+1}{(-2)}^r$

$=\frac{-1}{2\left(n+1\right)}{\sum }_{r=0}^n{ }^{n+1}{C}_{r+1}{(-2)}^{r+1}$

$=\frac{-1}{2\left(\text{n}+1\right)}\left[{(1-2)}^{\text{n}+1}-1\right]$

${\text{⇒P}}_{\text{n}}=\frac{1}{2\left(\text{n}+1\right)}\left[1-{(-1)}^{\text{n}+1}\right]$

$P_{2n}=\frac{1}{2\left(2n+1\right)}\left[1-{(-1)}^{2n+1}\right]$

$P_{2n}=\frac{1}{2n+1}$

\(\sum_{n=1}^{25} \frac{1}{P_{2 n}}=\sum_{n=1}^{25}(2 n+1)=3+5+\ldots . .+51=\frac{25}{2}[51+3]=25 \times 27=675\)

$$\begin{gathered} (1+x{)}^p(1-x{)}^q \\ =\left({}^{p}C_0+{}^{p}C_{1}x+{}^{p}C_2{x}^2+\dots \right) \\ \left({}^{q}C_0-{}^{q}C_{1}x+{}^{q}C_2{x}^2+\dots \right) \end{gathered}$$

$$\begin{gathered} \text{coefficient of} \mathrm{x} \text{is} \\ {}^{\mathrm{p}}\mathrm{C}_0{}^q\mathrm{C}_1-{}^{\mathrm{p}}\mathrm{C}_1{}^q\mathrm{C}_0=1 \end{gathered}$$

$$\begin{gathered} p-q=1 \\ \text{ coefficient of }{\mathrm{x}}^2 \mathrm{is} \\ {}^p\mathrm{C}_0{}^q\mathrm{C}_2-{}^p\mathrm{C}_1{}^q\mathrm{C}_1+{}^p\mathrm{C}_2{}^{\mathrm{q}}\mathrm{C}_0=-2 \\ \frac{q(q-1)}{2}-pq+\frac{p(p-1)}{2}=-2 \end{gathered}$$

$$\begin{gathered} q^2-q-2pq+p^2-p=-4 \\ 1-(p+q)=-4 [\text{since} p-q=1 ] \\ p+q=5 \\ p=3 , q=2 \\ \text{ so }{\mathrm{p}}^2+{\mathrm{q}}^2=13 \end{gathered}$$

$$\begin{gathered} 3^4\equiv 12(mod 23) \\ {3}^8\equiv 6(mod 23) \\ {3}^{11}\equiv 1(mod 23) \\ {3}^{99}\equiv 1(mod 23) \\ {3}^{99}.3^8\equiv 6(mod 23) \\ {3}^{107}\equiv 6(mod 23) \\ \mathrm{Hence} \mathrm{remainder} \mathrm{will} \mathrm{be} 6 \end{gathered}$$

Statement I:

If \(n\) is odd,

\(x^n+y^n=(x+y)\left(x^{n-1}-x^{n-2} y+x^{n-3} y^2-\cdots-x y^{n-2}+y^{n-1}\right)\) So,

\(x^n+y^n\) is divisible by \(x+y\).

\(25^{13}+3^{13}\) is divisible by \(25+3=28\).

Similarly :
\(20^{13}+8^{13}\) is divisible by \(20+8=28\).

So, \(25^{13}+20^{13}+8^{13}+3^{13}\) is divisible by \(7\). Statement I is true.

Statement II:

$R={\left(7+4\sqrt{3}\right)}^{25}=I+f$

$R^{\text{'}}={\left(7-4\sqrt{3}\right)}^{25}=f^{\text{'}}$

$\therefore R+R^{\text{'}}=2\left[{}^{25}{C}_0{7}^{25}+{}^{25}{C}_2{7}^{23}{\left(4\sqrt{3}\right)}^2+\mathrm{...}\right]$

\(l+f+f^{\prime}=\) even integer

Since \(I\) is an integer, \(f+f^{\prime}\) must also be an integer. Given \(0<f<1\) and \(0<f^{\prime}<1\), the only possible integer value for their sum is \(f+f^{\prime}=1\).

\(I=\) Even Integer \(-1=\) Odd Integer

The integral part I is indeed an odd number. Statement II is True.

⇒ Both the statement are correct

\( \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5 \)
\( =2\left\{{ }^5 \mathrm{C}_0 \cdot x^5+{ }^5 \mathrm{C}_2 \cdot x^3\left(x^3-1\right)+{ }^5 \mathrm{C}_4 \cdot x\left(x^3-1\right)^2\right\}\)
\( =2\left\{5 x^7+10 x^6+x^5-10 x^4-10 x^3+5 x\right\} \)
\( \Rightarrow \alpha=10, \beta=2, \gamma=-20, \delta=10\)

Now, \(1 0 u+2 v=1 8\)

\(-20 u+10 v=20\)
\(\Rightarrow u=1, v=4 \)
\(u+v=5\)

\(A=\) sum of coefficients in the expansion of \(\left(1-3 x+10 x^2\right)^n\)

put \(x=1\), we get \(A=8^n\)

\(B=\) sum of coefficients in the expansion of \(\left(1+x^2\right)^n\)

put \(x=1\), we get \(B=2^n\)

\((B)^3=8^n=A\)

\(A=\) sum of coefficients in the expansion of \(\left(1-3 x+10 x^2\right)^n\)

put \(x=1\), we get \(A=8^n\)

\(B=\) sum of coefficients in the expansion of \(\left(1+x^2\right)^n\)

put \(x=1\), we get \(B=2^n\)

\((B)^3=8^n=A\)

Required $= 1^{2} \cdot {}^{15}C_1 + 2^{2} \cdot {}^{15}C_2 + 3^{2} \cdot {}^{15}C_3 + \ldots + 15^{2} \cdot {}^{15}C_{15}$

$$\begin{array}{l} = \sum_{r=1}^{15} r^2 \cdot {}^{15}C_r \\ S = \sum_{r=1}^{15} r^2 \cdot \frac{15}{r} \cdot {}^{14}C_{r-1} = 15 \sum_{r=1}^{15} r^{14}C_{r-1} = 15 \sum_{r=1}^{15} (r - 1 + 1)^{14}C_{r-1} \\ = 15 \left[ \sum_{r=1}^{15} r - 1 \cdot \frac{14}{r-1} \cdot {}^{13}C_{r-2} + \sum_{r=1}^{15} {}^{14}C_{r-1} \right] \\ = 15 \times 14.2^{13} + 15.2^{14} \\ = 15.2^{13}(16) = 2^{17} \cdot 3^{1} \cdot 5^{1} \\ \end{array}$$

Then $m=17,n=1,k=1\Rightarrow m+n+k=19$

let \(N=\left((64)^{(64)}\right)^{64}\)

\(=(64)^{64^2}\)

\(N=(1+63)^{64^2}\)

let \(64^2=m\)

\(N=(1+63)^m\)

\(=1+{ }^m \mathrm{C}_1 63+{ }^m \mathrm{C}_2 (63)^2+\cdots+{ }^m \mathrm{C}_m (63)^m\)

\(=1+63 \mathrm{k}\)

Hence remainder is 1

$$\begin{gathered} {\left(\frac{1}{3}{x}^{\frac{1}{2}}+x^{-\frac{1}{4}}\right)}^{10} \\ {\text{general term T}}_{r+1}\text{ = }{}^{n}C_r{a}^r{b}^{10-r} \\ {T}_{r+1} ={}^{10}C_r{\left(\frac{1}{3}{x}^{\frac{1}{2}}\right)}^r{\left(x^{-\frac{1}{4}}\right)}^{10-r} \\ {T}_{r+1} ={}^{10}C_r\frac{1}{3^r}{x}^{\frac{r}{2}-\left(\frac{10-r}{4}\right)} \\ {T}_{r+1} ={}^{10}C_r\frac{1}{3^r}{x}^{\frac{2r-10+r}{4}} \\ {T}_{r+1} ={}^{10}C_r\frac{1}{3^r}{x}^{\frac{3r-10}{4}} ............\left(i\right) \\ \text{Now for} x^2 \\ \frac{3r-10}{4}=2 \\ r=6 \\ \text{put the value of r in eqn }\left(i\right), \text{we get} \\ {T}_7={}^{10}C_6.\frac{1}{3^6}.x^2 \\ {T}_7=\frac{210}{3^6}{x}^2 \\ {T}_7=\frac{70}{243}{x}^2 \end{gathered}$$

Using the standard binomial identity

$\sum_{r=1}^{n} r \binom{n}{r} x^r = nx(1 + x)^{n-1}$,

we substitute $n = 9$ and $x = \frac{1}{2}$:

$S_1 = 9 \cdot \left(\frac{1}{2}\right) \cdot \left(1 + \frac{1}{2}\right)^{9-1}$

$S_1=\frac{9}{2}\cdot {\left(\frac{3}{2}\right)}^8$

To express this in terms of $(\frac{3}{2})^9$,

we can rewrite it as: $S_1 = \frac{3^2}{2} \cdot \frac{3^8}{2^8} = \frac{3^{10}}{2^9} = 3 \cdot \frac{3^9}{2^9} = 3 \left(\frac{3}{2}\right)^9$

${\sum }_{r=1}^9\left(\genfrac{}{}{0px}{}{9}{r}\right){\left(\frac{1}{2}\right)}^r=\left[{\sum }_{r=0}^9\left(\genfrac{}{}{0px}{}{9}{r}\right){\left(\frac{1}{2}\right)}^r\right]-\left(\genfrac{}{}{0px}{}{9}{0}\right){\left(\frac{1}{2}\right)}^0$$={(1+\frac{1}{2})}^9-1={\left(\frac{3}{2}\right)}^9-1$

Therefore, $S_2 = 3 \left[ \left(\frac{3}{2}\right)^9 - 1 \right] = 3 \left(\frac{3}{2}\right)^9 - 3$

$S = S_1 + S_2$

$S = 3 \left(\frac{3}{2}\right)^9 + 3 \left(\frac{3}{2}\right)^9 - 3$

$S = 6 \left(\frac{3}{2}\right)^9 - 3$

$S = \alpha \left(\frac{3}{2}\right)^9 - \beta$

$(\alpha +\beta {)}^2=(6+3{)}^2=9^2=81$

\(
\begin{aligned}
& (1-x)\left[(1-x)\left(1+x+x^2\right)\right]^{2007} \\
& =(1-x)\left(1-x^3\right)^{2007} \\
& =\left(1-x^3\right)^{2007}-x\left(1-x^3\right)^{2007}
\end{aligned}
\)

\([(1- \left.x^3\right)^{2007}\) contains \(3 \lambda\) types of exponents while \(x\left(1-x^3\right)^{2007}\) will have \((3 \lambda+1)\) type while 2012 is
\((3 \lambda+2)\) type] that is not possible \(\Rightarrow 0\)
Coefficient of \(x^{2012}\) in \(\left(1-x^3\right)^{2007}=0\)
Coefficient of \(x^{2011}\) in \(\left(1-x^3\right)^{2007}=0\)
\(\Rightarrow\) Coefficient of \(x^{2012}\) in \((1-x)^{2008}\left(1+x+x^2\right)^{2007}=0\)

$$\begin{gathered} Given, \\ {\text{In (1+x)}}^{n+4} \text{coefficient of 5th, 6th \& 7th terms are in A.P.} \\ {}^{\mathrm{n}+4}\mathrm{C}_4,{}^{\mathrm{n}+4}\mathrm{C}_5,{}^{\mathrm{n}+4}\mathrm{C}_6, \text{are in A.P.} \\ \text{then ⇒}2\times {}^{\mathrm{n}+4}\mathrm{C}_5={}^{\mathrm{n}+4}\mathrm{C}_4+{}^{\mathrm{n}+4}\mathrm{C}_6 \\ \Rightarrow 4\times {}^{\mathrm{n}+4}\mathrm{C}_5=\left({}^{\mathrm{n}+4}\mathrm{C}_4+{}^{\mathrm{n}+4}\mathrm{C}_5\right)+\left({}^{\mathrm{n}+4}\mathrm{C}_5+{}^{\mathrm{n}+4}\mathrm{C}_6\right) \\ \Rightarrow 4\times {}^{\mathrm{n}+4}\mathrm{C}_5={}^{\mathrm{n}+5}\mathrm{C}_5+{}^{\mathrm{n}+5}\mathrm{C}_6 \\ \Rightarrow 4\times \frac{(\mathrm{n}+4)!}{5!·(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!·\mathrm{n}!} \\ \Rightarrow {\mathrm{n}}^2+11\mathrm{n}+30=24\mathrm{n} \\ \Rightarrow \mathrm{n}=3,10\text{ (rejected) } \\ \therefore \text{ Largest binomial coefficient} \\ \text{in expansion of }(1+\mathrm{x}{)}^7 \text{ is coeff.} \\ \text{o}\text{f middle term } \\ {\Rightarrow }^7{\mathrm{C}}_4={}^7\mathrm{C}_3=35 \end{gathered}$$

$(1+x{)}^{2n-1}$
$T_{30}={}^{2n-1}C_{29}{x}^{29}$
$T_{12}={}^{2n-1}C_{11}{x}^{11}$
$A={}^{2n-1}C_{29}$
$B={}^{2n-1}C_{11}$
$2\times \frac{(2n-1)!}{29!\times (2n-30)!}=\frac{5\times (2n-1)!}{11!\times (2n-12)!}$
$$\begin{gathered} \frac{2\times 30}{30!\times (2n-30)!}=\frac{5}{11!\times (2n-12)!} \\ \Rightarrow 12!\times (2n-12)!=30!\times (2n-30)! \end{gathered}$$
Comparing

$2n-12=30\Rightarrow n=21$ or $2n-30=12 \Rightarrow n=21$

$$\begin{gathered} {\left(x+\sqrt{x^3-1}\right)}^5+{\left(x-\sqrt{x^3-1}\right)}^5 \\ =2\left\{{}^5\mathrm{C}_0·{\mathrm{x}}^5+{}^5\mathrm{C}_2·{\mathrm{x}}^3\left({\mathrm{x}}^3-1\right)+{}^5\mathrm{C}_4·\mathrm{x}{\left({\mathrm{x}}^3-1\right)}^2\right\} \\ =2\left\{5x^7+10x^6+x^5-10x^4-10x^3+5x\right\} \\ \Rightarrow \alpha =10,\beta =2,\gamma =-20,\delta =10 \\ \text{ Now, } \\ 10\mathrm{u}+2\mathrm{v}=18 \\ \Rightarrow 5\mathrm{u}+\mathrm{v}=9 ...\left(1\right) \\ -20 u+10 v=20 \\ \Rightarrow -2u+v=2 ...\left(2\right) \\ u=1 and v=4 \\ \Rightarrow u+v=5 \end{gathered}$$

$$\begin{gathered} S=\sum _{r=50}^{100}\frac{{}^{100}C_r}{r+1}=\sum _{r=50}^{100}\frac{1}{r+1}·\frac{r+1}{101}·{}^{101}C_{r+1} \\ S=\frac{1}{101}\sum _{r=50}^{100}{}^{101}C_{r+1} \\ \because \sum _{r=0}^n{}^{n}C_r=2^n , {}^{n}C_r={}^{n}C_{n-r} \\ \Rightarrow \sum _{r=0}^{50}{}^{101}C_r=\sum _{r=51}^{101}{}^{101}C_r=\frac{2^{101}}{2}=2^{100} \\ \therefore S=\frac{1}{101}\times \frac{2^{101}}{2}=\frac{2^{100}}{101} \end{gathered}$$

$(1+x{)}^{2n-1}$
$T_{30}={}^{2n-1}C_{29}{x}^{29}$
$T_{12}={}^{2n-1}C_{11}{x}^{11}$
$A={}^{2n-1}C_{29}$
$B={}^{2n-1}C_{11}$
$2\times \frac{(2n-1)!}{29!\times (2n-30)!}=\frac{5\times (2n-1)!}{11!\times (2n-12)!}$
$$\begin{gathered} \frac{2\times 30}{30!\times (2n-30)!}=\frac{5}{11!\times (2n-12)!} \\ \Rightarrow 12!\times (2n-12)!=30!\times (2n-30)! \end{gathered}$$
Comparing

$2n-12=30\Rightarrow n=21$ or $2n-30=12 \Rightarrow n=21$

$$\begin{gathered} \text{let coefficients of} {\mathrm{T}}_{\mathrm{r}},{ \mathrm{T}}_{\mathrm{r}+1},{ \mathrm{T}}_{\mathrm{r}+2} \\ \text{in the expansion of} (a+b{)}^{12} \text{are in GP} \\ \text{ So, }\frac{T_{r+1}}{T_r}=\frac{T_{r+2}}{T_{r+1}} \\ \frac{{}^{12}\mathrm{C}_{\mathrm{r}}}{{}^{12}\mathrm{C}_{\mathrm{r}-1}}=\frac{{}^{12}\mathrm{C}_{\mathrm{r}+1}}{{}^{12}\mathrm{C}_{\mathrm{r}}} \\ \frac{12-r+1}{r}=\frac{12-(r+1)+1}{r+1} \\ \left(13 – r\right)\left(r + 1\right) = \left(12 –r\right)\left(r\right) \\ 13r+13-r^2-r=12r-r^2 \\ 12r+13=12r \\ 13=0 \\ \text{No value of r possible} \\ \text{ So} p = 0 \\ {\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)}^{12}=\sum _{r=1}^{12}{}^{12}C_r{\left(3^{\frac{1}{4}}\right)}^{12-r}{\left(4^{\frac{1}{3}}\right)}^r \end{gathered}$$

$$\begin{gathered} \text{when} r=0, \text{rational term}=27 \\ \text{when} r=12, \text{rational term}=256 \end{gathered}$$

$$\begin{gathered} \mathrm{q} = 27 + 256 = 283 \\ \mathrm{p} + \mathrm{q} = 0 + 283 = 283 \end{gathered}$$

We have \(\left(5^{\frac{1}{2}}+7^{\frac{1}{8}}\right)^{1016}\)

${(5^{\frac{1}{2}}+7^{\frac{1}{8}})}^{1016}={\sum }_{r=0}^{1016} {}^{1016}{C}_r{5}^{\frac{1016-r}{2}}{7}^{\frac{r}{8}}$

For integral terms, \(r\) should be multiple of 8

\(\therefore r \in\{0,8,16, \ldots \ldots, 1016\}\)

No. of integral terms = 128

Using Vandermonde’s identity:

\({ }^3 \mathrm{C}_3 \cdot{ }^{30} \mathrm{C}_{30-\mathrm{r}}+{ }^3 \mathrm{C}_2 \cdot{ }^{30} \mathrm{C}_{31-\mathrm{r}}+{ }^3 \mathrm{C}_1 \cdot{ }^{30} \mathrm{C}_{32-\mathrm{r}}+{ }^3 \mathrm{C}_0 \cdot{ }^{30} \mathrm{C}_{33-\mathrm{r}}\)

\(={ }^{33} C_{33-r}={ }^{33} C_r\),

so, \({ }^m C_r={ }^{33} C_r\)

\(\Rightarrow m=33\)