Question

Find the sum of series: 1^2+2+3^2+4+5^2+6+..... to 2n terms

Answer 1

Answer:

a= 1

n=2n

d = a2-a1

= 2-1

= 1

Sn= n/2 ( 2a +(n-1) d)

=2n /2 (2×1+ (2n-1 ) 1)

= n (2+ 2n-1 )

= n (2n +1 )

= 2n2 +n