Question

please can anyone solve it​

Answer 1

Given,

$\small\displaystyle\longrightarrow\sum_{k=1}^N\dfrac{2k+1}{(k^2+k)^2}=0.9999$

We take,

• $\small2k+1=k^2+2k+1-k^2=(k+1)^2-k^2$

• $\small(k^2+k)^2=\big(k(k+1)\big)^2=k^2(k+1)^2$

Then,

$\small\displaystyle\longrightarrow\sum_{k=1}^N\dfrac{(k+1)^2-k^2}{k^2(k+1)^2}=0.9999$

$\small\displaystyle\longrightarrow\sum_{k=1}^N\left(\dfrac{1}{k^2}-\dfrac{1}{(k+1)^2}\right)=0.9999$

$\small\displaystyle\longrightarrow\sum_{k=1}^N\dfrac{1}{k^2}-\sum_{k=1}^N\dfrac{1}{(k+1)^2}=0.9999$

We see that,

• $\small\displaystyle\sum_{k=a}^bf(k)=\sum_{k=a+r}^{b+r}f(k-r)$

Thus second sum in the LHS can be taken as, [a = 1, b = N, f(k) = 1/[(k+1)^2], r = 1]

• $\small\displaystyle\sum_{k=1}^N\dfrac{1}{(k+1)^2}=\sum_{k=2}^{N+1}\dfrac{1}{k^2}$

Then,

$\small\displaystyle\longrightarrow\sum_{k=1}^N\dfrac{1}{k^2}-\sum_{k=2}^{N+1}\dfrac{1}{k^2}=0.9999$

The term for k=1 is taken out from first sum, and the term for k=N+1 is taken out from second sum.

$\small\displaystyle\longrightarrow\left(\dfrac{1}{1^2}+\sum_{k=2}^N\dfrac{1}{k^2}\right)-\left(\sum_{k=2}^N\dfrac{1}{k^2}+\dfrac{1}{(N+1)^2}\right)=0.9999$

$\small\displaystyle\longrightarrow1+\sum_{k=2}^N\dfrac{1}{k^2}-\sum_{k=2}^N\dfrac{1}{k^2}-\dfrac{1}{(N+1)^2}=0.9999$

$\small\displaystyle\longrightarrow1-\dfrac{1}{(N+1)^2}=\dfrac{9999}{10000}$

$\small\displaystyle\longrightarrow\dfrac{1}{(N+1)^2}=1-\dfrac{9999}{10000}$

$\small\displaystyle\longrightarrow\dfrac{1}{(N+1)^2}=\dfrac{1}{10000}$

$\small\displaystyle\longrightarrow(N+1)^2=10000$

$\small\displaystyle\longrightarrow N+1=\pm100$

$\small\displaystyle\longrightarrow N=\pm100-1$

$\small\displaystyle\longrightarrow N=99\quad OR\quad N=-101$

Since N is a positive integer,

$\small\text{ \displaystyle\longrightarrow\underline{\underline{N=99}} }$

Hence the value of N is 99.