The sum 1+ 12 (12+22 )+ 13 (12+22+32 )+ upto 10 terms is equal to :
- 130
- 155
- 3152
- 3252
Answer
(3) 3152
Solution
The general term is Tn = 1n (12 + 22 + + n2 ) = 1n n(n + 1)(2n + 1)6 = 2n2 + 3n + 16, so
S10 = 16 [2 n = 110n2 + 3 n = 110n + 10 ]
S10 = 16 [2(385) + 3(55) + 10 ]
S10 = 9456 = 3152
Answer: (3)