InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Find the sum to n terms of the series, whose `n^(t h)`terms is given by : `(2n-1)^2` |
|
Answer» Here, `a_n = (2n-1)^2 = 4n^2+1-4n` So, `S_n = sum_(n=1)^na_n = sum_(n=1)^n4n^2+sum_(n=1)^n1-sum_(n=1)^n4n` `S_n = 4((n(n+1)(2n+1))/6)+n-4((n(n+1))/2)` `=2/3(n(n+1)(2n+1))+n-2(n(n+1))` `=2n(n+1)((2n+1)/3-1)+n` `=2n(n+1)((2n-2)/3)+n` `=4/3n(n+1)(n-1)+n` `=n(4/3(n^2-1)+1)` `=n((4n^2-1)/3)` `:. S_n = n/3(4n^2-1)` |
|