Question

the sum of n terms of the series
1²/1 + (1²+2²)/2 + (1²+2²+3²)/3.....is
A) (n/36)(4n²+15n+17) B)(n/6)(n+1)(n-1)
c) (n²+3n+20) D) none

Answer 1

testing for 3 terms, (n=3)

1²/1 + (1²+2²)/2 + (1²+2²+3²)/3
= 1 + 5/2 + 14/3
= 49/6

for n = 3
a) --> (3/36)(36+45+17) = 49/6
b) --> (3/6)((4)(2) = 4 , noway!
c) --> (9+9+29) = 47 , noway!

suspicious it could be a)
test for n = 2
sum(2) = 1 + 5/2 = 7/2
a) --> (2/36)(16 + 30 +17) = 7/2

testing for n = 4
sum4 = 1+5/2+14/3+ <b>(1^2+2^2+3^2+4^2)/4</b> = 15/2
using a) --> (4/36)(64 + 60 + 17) = 47/3 ≠ 15/2

so it has to be none of these or D)

Answer 2

Ok