Question

Find the following sum: 1/(2^2 –1) +1/(4^2 –1) + 1/(6^2 –1) + …. +1/(20^2 –1)​

Answer 1

Answer:

n these summations , we first write the nth term

Notice here the nth term will be  1/(n)(n+1)  

It can be written as  (n+1)−n/(n)(n+1)  

=>1/n−1/(n+1)  

Now when you take summations from n=1 to infinity, you can observe cancellation of terms

T1  =1/1−1/2  

T2  =1/2−1/3  

T3  =1/3−1/4  

And so on till infinity

As by calculation we will be left with 1 -  1/infinity  the negative term will be approximately zero therefore the answer will be 1

Hope this helps!

22.4K viewsView 20 Upvoters · Answer requested by Jagan Subramanian

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Find the value of (1/(2*3)) + (1/(3*4)) + (1/(4*5)) + (1/(5*6)) +…+ ((1/ (9*10))?

How do I solve these types of problems:  11⋅2+12⋅3+13⋅4+...+199⋅100 ?

How do I find the sum of the series:  11⋅2+12⋅3+13⋅4+14⋅5+…+1n(n+1)?

Step-by-step explanation:

Answer 2

Answer:

The sum of the series is $\frac{10}{21}$

Step-by-step explanation:

This is a series whose terms has some pattern.

Pattern suggests that, $n$th term is $\frac{1}{(2 n)^{2}-1}$

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

Now we see that,$k$ th term $=\frac{1}{2}\left(\frac{1}{(2 k-1)}-\frac{1}{(2 k+1)}\right)$

And $(k+1)$ th term is $\frac{1}{2}\left(\frac{1}{(2 k+1)}-\frac{1}{(2 k+3)}\right)$

Adding these consecutive terms, cancels out, $\frac{1}{(2 k+1)}$. Similarly, when we add all these terms of series, all terms sum to zero, except $\frac{1}{1} and \frac{1}{21}$. This is the main insight of the solution.

$\sum_{n=1}^{10} \frac{1}{2 n^{2}-1}=\sum_{n=1}^{10} \frac{1}{2}\left(\frac{1}{(2 n-1)}-\frac{1}{(2 n+1)}\right)\\=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{21}\right)=\frac{10}{21}$

$\frac{1}{2^{2}-1}+\frac{1}{4^{2}-1}+\frac{1}{6^{2}-1}+\cdots+\frac{1}{20^{2}-1}=\frac{10}{21}$