Question

find the sum of first n terms of the series 1²+2²+....+n²​

Answer 1

Answer:

a1=1,a2=4....n square

d=4-1=3

an=2a+(n-1)d

an=2(1)+(n-1)3

an=2+3n-3

an=-1+3n

an=3n-1

sn=n/2(a+an)

sn=n/2(1+3n-1)

sn=n/2(3n)

sn=3n square /2