If \(2{a}_{n+2}=5{a}_{n+1}-3{a}_{n}\), where \(n=0,1,2,\ldots\). If \({a}_{0}=\)3 and \({a}_{1}=4\), then the value of \…

\(2{a}_{n+2}=5{a}_{n+1}-3{a}_{n}\\ \Rightarrow 2{t}^{2}-5t+3=0\\ \Rightarrow t=1,\frac{3}{2}\\ {a}_{n}=A\cdot (1{)}^{n}+B\cdot {\left(\frac{3}{2}\right)}^{n}\\ {a}_{0}=3,{a}_{1}=4\\ thenA=1&B=2\\ Now,{a}_{n}=1+2{\left(\frac{3}{2}\right)}^{n}\\ {a}_{100}=1+2{\left(\frac{3}{2}\right)}^{100}\\ {\left(\frac{3}{2}\right)}^{100}=\frac{{a}_{100}-1}{2}\\ Now\sum _{k=1}^{100}{a}_{k}={S}_{100}=100+6\left({\left(\frac{3}{2}\right)}^{100}-1\right)\\ =100+6\left(\frac{{a}_{100}-1}{2}-1\right)\\ =3{a}_{100}+91\)