1^2 +2.2^2+3^3+2.2^4+5^2+5^2+2.6^2 ........ is n(n+1)^2/2 when "n" is even . What is the sum when n is odd?

Asked by  | 4th Nov, 2013, 11:03: PM

Expert Answer:

When n is odd, last term will be n^2.
 
Thus, 
Sum = 1^2 +2.2^2+3^3+2.2^4+5^2+2.6^2 ... + 2.(n-1)^2 + n^2
= [(n-1)(n)^2]/2 + n^2      [Replacing n by (n - 1)]
= [n^3 + n^2]/ 2 = [n^2(n+1)]/2

Answered by  | 4th Nov, 2013, 11:29: PM

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