Question

Pls solve for the following:​

Answer 1

Answer:

$\huge{\orange{Hey\:Mate}}$

To find the sum of 1/(1*2) +1/(2*3)+1/(3*4)+....1/n(n+1).

Let Sn = 1/(1*2) +1/(2*3)+1/(3*4)+....1/n(n+1).

Consider 1/n(n+1) = 1/n -1/(n+1)

Threfore we can split each term of the given series as below:

Sn = {(1/1-1/2) + (1/2 -1/3) +(1/3-1/4) + (1/4-1/5) +..... +(1/n-1/(n+1)}

Sn = 1-1/2 +1/2 -1/3 +1/3 -1/4 +1/4 +......-/n+1/n -1/(n+1).

Sn = 1 +0 +0+...+0-1/(n+1)

Therefore Sn = 1-1/(n+1)

Or Sn = (n+1-1)/(n+10.

Sn = n/(n+1).

Therefore the sum to n terms of 1/(1*2) +1/(2*3)+1/(3.4)+....1/n(n+1) is equal to

=n/(n+1) ${\red{Ans}}$

$\large{\orange{Hope\:it\:helped}}$

Answer 2

Answer:

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