Find the sum of the first 40 terms of the series 1² – 2² + 3² – 4² + … .

Let the sum of n terms be,

S_n= 1² –2² +3² –4²+5² –6²+7² –8² ….

= (1² –2²) + (3² –4²) + (5² –6²) + (7²–8²) …..

= (1–4) + (9–16) + (25–36) + (49–64)……..

= –3 –7 –11 –15………………. [No. of terms ]

This represents an A.P.

(NOTE: Here the number of terms has been halved. If n= no. of terms in original series and m= no. of terms in new A.P. then )

Now, here a= –3, d= (–7 – (–3)) = –4

Sum of m terms =

=

=

=

=

=

As n= 40,

Required sum =

= –820