# Question 8d376

Feb 6, 2017

Refer to the Explanation given below for the Answer : $2017 - \frac{1}{2017.}$

#### Explanation:

Let us observe that the general ${n}^{t h} \text{ term } {T}_{n}$ of the given

Series is given by,

${T}_{n}^{2} = 1 + \frac{1}{n} ^ 2 + \frac{1}{n + 1} ^ 2$

$= \frac{{n}^{2} {\left(n + 1\right)}^{2} + {\left(n + 1\right)}^{2} + {n}^{2}}{{n}^{2} {\left(n + 1\right)}^{2}}$

$= \frac{{\left\{n \left(n + 1\right)\right\}}^{2} + {n}^{2} + 2 n + 1 + {n}^{2}}{n \left(n + 1\right)} ^ 2$

$= \frac{{\left\{n \left(n + 1\right)\right\}}^{2} + 2 {n}^{2} + 2 n + 1}{n \left(n + 1\right)} ^ 2$

$= \frac{{\left\{n \left(n + 1\right)\right\}}^{2} + 2 \left(1\right) \left\{n \left(n + 1\right)\right\} + {1}^{2}}{{\left\{n \left(n + 1\right)\right\}}^{2}}$

$= {\left\{n \left(n + 1\right) + 1\right\}}^{2} / \left[{\left\{n \left(n + 1\right)\right\}}^{2}\right\}$

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

$\Rightarrow {T}_{n} = \frac{n \left(n + 1\right) + 1}{n \left(n + 1\right)} = \frac{n \left(n + 1\right)}{n \left(n + 1\right)} + \frac{1}{n \left(n + 1\right)}$

$\Rightarrow {T}_{n} = 1 + \frac{\left(n + 1\right) - n}{n \left(n + 1\right)} = 1 + \frac{\cancel{n + 1}}{n \left(\cancel{n + 1}\right)} - \frac{\cancel{n}}{\cancel{n} \left(n + 1\right)}$

$\therefore {T}_{n} = 1 + \frac{1}{n} - \frac{1}{n + 1}$

Hence, ${S}_{n} = \sum {T}_{n} = \sum \left[1 + \frac{1}{n} - \frac{1}{n + 1}\right]$

$= \sum 1 + \sum \left\{\frac{1}{n} - \frac{1}{n + 1}\right\} = n + \sum \left\{\frac{1}{n} - \frac{1}{n + 1}\right\}$

=n+{(1/1cancel(-1/2))+(cancel(1/2)cancel(-1/3))+(cancel(1/3)cancel(-1/4))+...+(cancel(1/(n-1))cancel(-1/n))+(cancel(1/n)-1/(n+1)}#

$\text{So, in general, } {S}_{n} = n + \left\{1 - \frac{1}{n + 1}\right\} = \left(n + 1\right) - \frac{1}{\left(n + 1\right)} .$

In particular, ${S}_{2016} = 2017 - \frac{1}{2017.}$

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