Let a_1=1 and for n 1,a_n+1=(1)/(2)a_n+n^2-2n-1n^2(n+1)^2.

Question

Let ${a}_{1}=1$ and for $n\geqslant 1,{a}_{n+1}=\frac{1}{2}{a}_{n}+\frac{{n}^{2}-2n-1}{{n}^{2}(n+1{)}^{2}}$. Then $|\sum _{\infty}^{n=1}({a}_{n}-\frac{2}{{n}^{2}})|$ is equal to $____$ .

Answer: 2

Accepted Answer

$\begin{gathered} {a}_{n+1}=\frac{1}{2}{a}_{n}+\frac{1}{(n+1{)}^{2}}-\frac{((n+1{)}^{2}-{n}^{2})}{{n}^{2}(n+1{)}^{2}} \ {a}_{n+1}=\frac{{a}_{n}}{2}+\frac{2}{(n+1{)}^{2}}-\frac{1}{{n}^{2}} \ {a}_{n+1}-\frac{2}{(n+1{)}^{2}}=\frac{1}{2}({a}_{n}-\frac{2}{{n}^{2}}) \ {b}_{n}={a}_{n}\frac{-2}{{n}^{2}} \end{gathered}$

$\begin{gathered} \langle bangle =\frac{1}{2} \ {b}_{1}={a}_{1}-\frac{2}{1}=1-2=(-1) \ \sum _{n=1}^{\infty}({a}_{n}-\frac{2}{{n}^{2}})=\sum _{n=1}^{\infty}{b}_{n}=\frac{(-1)}{1-(\frac{1}{2})}=\frac{-1}{1/2}=-2 \ \Rightarrow |\sum _{n=1}^{\infty}({a}_{n}-\frac{2}{{n}^{2}})|=2 \end{gathered}$

Solutions · 1