IIT JEE , AIEEE Forum - Mathematics , Algebra

harshiiit

3 + 5 + 7 + ............................n

------------ ------ --------- 1^2*2^2 2^2*3^2 3^2*4^2

Avinash_Bhat

I think your question is :

3 / 1²*2² + 5 / 2²*3² + 7 / 3²*4² + ......................... upto n terms.

Here, nth term : Tn = (2n + 1) / n²(n + 1)²

= ( 1/n(n + 1) ) ( (2n + 1) / n(n + 1) )

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

= ( 1/n² - 1/(n + 1)² )

SO : Sum of all the n terms : Sn = Tn = 1/1² - 1/(n + 1)²

= 1 - 1/(n + 1)²

shine

see i have a trick.....................u might b knowing tht
if a& b r 2 consecutive no's then a^2 -b^2 =a+b
same is here
see
3 / 1^2*2^2 + 5 / 2^2*3^2 + 7 / 3^2*4^2 + ......................... upto n terms.
can b written as
2^2-1^2/1^2*2^2 +3^2-2^2/ 2^2*3^2 + 4^2-3^2/ 3^2*4^2 ...............n^2-(n-1)^2/n+1^2*(n)^2
1-1/4 +1/4 -1/9 +1/9 /........................-1/(n+1)^2
after cancelling we get
1 - 1/(n + 1)2

hope that was easy

edison

Let,

S = 3 / 1²*2² + 5 / 2²*3² + 7 / 3²*4² + ... upto n terms.

Where nth term being = (2n + 1) / n²(n + 1)²

= [ 1/n(n + 1) ] [(2n + 1) / n(n + 1)]

splitting the two factors by partial fractions we have

= [1/n - 1/(n + 1) ] [ 1/n + 1/(n + 1)]

= [ 1/n² - 1/(n + 1)² ]

Hence Sum upto n terms is given by

S = [ 1/n² - 1/(n + 1)² ]

or S = 1/1² - 1/(n + 1)²(as other terms cancel)

or S = 1 - 1/(n + 1)²

as very well explained by avinash bhat

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