Question

if 1^2+2^2+3^2+...+n^2=3025,then n=

Answer 1

Answer:

Correct option is B)

Sum of cubes of 1st m natural numbers is given by:

4

m

2

(m+1)

2

3025=

4

m

2

(m+1)

2

12100=m

2

(m+1)

2

Taking square roots on both sides we get,

110=m(m+1)

m

2

+m−110=0

(m−10)(m+11)=0

m=10 or m=−11

Since m cannot be negative, m=10.