NIMCET 2013 Question Paper with solutions

Question : 49 of 120

Marks: +1, -0

Sum of 20 terms of the series -

1^2+2^2-3^2+4^2

- ...

Solution:

Given, series -

1^2+2^2-3^2+4^2+...

=\{(-1)^{2}+(2)^{2}\}+\{(-3)^{2}+(4)^{2}\}+. . .+\{(-19)^{2}+(20)^{2}\}

=3+7+11+. . .+39

Required sum (for 10 terms)

=({3+39}/2)×10=210

Go to Question:

Prev Question

Next Question