SOLUTION: the sum of n terms of 1²/1 + (1²+2²)/2 + (1²+2²+3²)/3+.......is a)(n/36)(4n²+15n+17) b) (n/36)(4n²-15n-17) c)(n/36)(4n²-15n+17) d) none

Algebra ->  Sequences-and-series -> SOLUTION: the sum of n terms of 1²/1 + (1²+2²)/2 + (1²+2²+3²)/3+.......is a)(n/36)(4n²+15n+17) b) (n/36)(4n²-15n-17) c)(n/36)(4n²-15n+17) d) none      Log On


   

Question 941377: the sum of n terms of 1²/1 + (1²+2²)/2 + (1²+2²+3²)/3+.......is
a)(n/36)(4n²+15n+17) b) (n/36)(4n²-15n-17) c)(n/36)(4n²-15n+17) d) none
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
the sum of n terms of 1²/1 + (1²+2²)/2 + (1²+2²+3²)/3+.......is
a)(n/36)(4n²+15n+17) b) (n/36)(4n²-15n-17) c)(n/36)(4n²-15n+17) d) none
sum%28%28expr%281%2Fk%29%2Asum%28%28i%5E2%29%2C1=1%2Ck%29%29%2Ck=1%2Cn%29

Substituting sum of first k squares: sum%28%28i%5E2%29%2C1=1%2Ck%29=k%28k%2B1%29%282k%2B1%29%2F6








sum%28%0D%0A%0D%0A%28%28k%2B1%29%282k%2B1%29%2F6%29%0D%0A%0D%0A%0D%0A%2Ck=1%2Cn%29





expr%281%2F6%29sum%28%0D%0A%0D%0A%282k%5E2%2B3k%2B1%29%0D%0A%0D%0A%0D%0A%2Ck=1%2Cn%29



Since sum%28%281%29%2Ck=1%2Cn%29=n, and factoring out constants,



Since sum%28%28k%5E2%29%2Ck=1%2Cn%29=n%28n%2B1%29%282n%2B1%29%2F6, and sum%28%28k%29%2C1=1%2Cn%29=n%28n%2B1%29%2F2







Getting an LCD



expr%281%2F6%29%28+%282n%28n%2B1%29%282n%2B1%29%2B9n%28n%2B1%29%2B6n%29%2F6%29%29

Factor out n in the top, multiplying the two 6 denominators:

+%28n%282%5E%22%22%28n%2B1%29%282n%2B1%29%2B9%28n%2B1%29%2B6%29%29%2F36%29

+expr%28n%2F36%29%282%5E%22%22%28n%2B1%29%282n%2B1%29%2B9%28n%2B1%29%2B6%29%29

+expr%28n%2F36%29%282%5E%22%22%282n%5E2%2B3n%2B1%29%2B9n%2B9%2B6%29%29

+expr%28n%2F36%29%284n%5E2%2B6n%2B2%2B9n%2B15%29

+expr%28n%2F36%29%284n%5E2%2B15n%2B17%29

Edwin