SOLUTION: sum to n terms of (1/1cube)+(1+2/1cube+2cube)+......... is
Algebra.Com
Question 486415: sum to n terms of (1/1cube)+(1+2/1cube+2cube)+......... is
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Use the fact that
^2} = \sum_{i=1}^n \frac{1}{1+2+...+i})
Since the sum of the first i positive integers is i(i+1)/2, we reciprocate so our sum is equal to
} = 2\sum_{i=1}^n \frac{1}{i(i+1)})
I learned a nice Mathcounts trick that you can decompose that fraction to 1/i - 1/(i+1) (if you don't know it, just decompose it normally using partial fractions). This becomes a "telescoping" sum where all the terms collapse except the first and last:
)
.
RELATED QUESTIONS
please,sir,help me urgently solve this question.
the question is- 1cube = 1square
(answered by 303795)
If there are (2n+1) terms in an arithmetic series, prove that the ratio of the sum of odd (answered by richard1234)
The sum
6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right)
is equal to a polynomial f(n)... (answered by Edwin McCravy)
The sum n terms is given by sn=n/2(1+n) find the fifth term of the... (answered by ewatrrr)
the sum to n terms of a sequence of numbers is given as Sn=n/2(5n+9).
1.calculate the... (answered by richwmiller)
The sum to n terms of a sequence of numbers is given as : Sn = n/2(5n+9)
1. Calculate... (answered by ikleyn)
Find the sum of n terms of a G.P. whose nth term is... (answered by amalm06,ikleyn)
The sum of a finite geometric sequence for n terms with common ratio r and initial term... (answered by solver91311,ewatrrr,richard1234)
Find the sum of the first 60 terms of the sequence below.
15,17,2,-15,-17
(Where {{{a... (answered by KMST)