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| 1. |
Find the sum of series (2^2+4^2+6^2+8^2+........to n terms |
| Answer» nth term of given series = an = square of (2n), Therefore, sn= a1+a2+a3+......+an= 4(sqr(1)+sqr(2)+sqr(3)+......sqr(n))Using identity , sum of square of n natural numbers = n(n+1)(2n+1)/6Sn= 4n(n+1)(2n+1)/6 | |