Find-the-nth-term-of-the-sequence-a-n-such-that-a-1-1-a-n-1-1-2-a-n-n-2-2n-1-n-2-n-1-2-n-1-2-3-

Question Number 98205 by I want to learn more last updated on 12/Jun/20

Find the nth term of the sequence {a_{n}} such that

a_{1} = 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n^{2} - 2 n - 1}{n^{2} {(n + 1)}^{2}} (n = 1, 2, 3, \dots)

Answered by mr W last updated on 12/Jun/20

\frac{n^{2} - 2 n - 1}{n^{2} {(n + 1)}^{2}}

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

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

\Rightarrow a_{n + 1} = \frac{1}{2} a_{n} + \frac{2}{{(n + 1)}^{2}} - \frac{1}{n^{2}}

\Rightarrow a_{n + 1} - \frac{2}{{(n + 1)}^{2}} = \frac{1}{2} (a_{n} - \frac{2}{n^{2}})

l e t b_{n} = a_{n} - \frac{2}{n^{2}}

\Rightarrow b_{n + 1} = \frac{1}{2} b_{n} \Rightarrow G . P .

\Rightarrow b_{n} = b_{1} {(\frac{1}{2})}^{n - 1} = (a_{1} - \frac{2}{1^{1}}) \frac{1}{2^{n - 1}} = - \frac{1}{2^{n - 1}}

\Rightarrow a_{n} = - \frac{1}{2^{n - 1}} + \frac{2}{n^{2}} = 2 (\frac{1}{n^{2}} - \frac{1}{2^{n}})

Commented by I want to learn more last updated on 12/Jun/20

Thanks sir

Commented by I want to learn more last updated on 12/Jun/20

Ok, each time, i will think how to transform it to linear .

Commented by Rio Michael last updated on 12/Jun/20

sir please any hint on how to approach Q 98208 ? ?