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Sn = 3/1^3 + 5/(1^3+2^3+3^3) + 7/(1^3+2^3+3^3).... upto n terms, then lim n - infinity Sn

Sn = 3/1^3 + 5/(1^3+2^3+3^3) + 7/(1^3+2^3+3^3).... upto n terms, then lim n - infinity Sn

Grade:12

4 Answers

Akash Kumar Dutta
98 Points
8 years ago

Dear Shuvam,
Ans is 4

Regards.

Shuvam Shukla
37 Points
8 years ago
Actually I need the process of doing..
Shivam Dimri
43 Points
8 years ago

its easy

general term is Tn = 4 * (2n+1)/(n2(n+1)2)

i hope you know the formula of 13 + 23 + 23 .... = n2(n+1)2/4

now split the general term as 

numerator = 2n+1 = (n+1) + (n) 

Tn = 4* [ 1/n2(n+1)   +   1/n(n+1)2]

again split the numerator

this time 1 = n+1 - n

=> T= 4* [ 1/n2  - 1/n(n+1)  +  1/n(n+1)   - 1/(n+1)2]

=> Tn = 4* [ 1/n - 1/(n+1)2]

if n-> infinity

then the subsequent terms gets cancelled and the last term tends to zero as n is in denominator

=> Sn = T1  + T2  +  T3  ......

=> Sn = 4*[1 - 1/4  + 1/4 .....]

=> Sn = 4

got it!!

Akash Kumar Dutta
98 Points
8 years ago

the general term is = 4(2n+1) /  [n(n+1)]^2
now apply partial fractions
and then add one by one...
you get the ans.
its that simple!

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