JEE Main2012MathematicsSequences and SeriesActual
If the sum of the series 1^2+2 2^2+3^2+2 4^2+5^2+ ... 2.6^2+ upto n terms, when n is even, is n(n+1)^2 2 , then the sum of the series, when n is odd, is
Options
- An^2(n+1)
- Bn^2(n-1) 2
- Cn^2(n+1) 2
- Dn^2(n-1)
Correct answer
C. n^2(n+1) 2
Step-by-step solution
If n is odd, the required sum is aligned & 1^2+2.2^2+3^2+2.4^2+ . .+2(n-1)^2+n^2 & = (n-1)(n-1+1)^2 2 +n^2 ( n-1 is even ) & = ( n-1 2 +1 ) n^2= n^2( n +1) 2 aligned