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Sum of series

412
2
1. Homework Statement

Find the sum of the series:

[tex]
S = 1~+~\frac{1 + 2^3}{1 + 2}~+~\frac{1 + 2^3 + 3^3}{1 + 2 + 3}~+~\frac{1 + 2^3 + 3^3 + 4^3}{1 + 2 + 3 + 4}~+~...
[/tex]

upto 11 terms

2. Homework Equations

Sum of first 'n' natural numbers: [tex]S = \frac{n(n + 1)}{2}[/tex]
Sum of the squares of the first 'n' natural numbers: [tex]S = \frac{n(n + 1)(2n + 1)}{6}[/tex]
Sum of the cubes of the first 'n' natural numbers: [tex]S = \left(\frac{n(n+1)}{2}\right)^2[/tex]


3. The Attempt at a Solution

In the series, the [itex]n^{th}[/itex] term is given by:

[tex]
T_n = \frac{1 + 2^3 + 3^3 + ... + n^3}{1 + 2 + 3 + ... + n}
[/tex]

[tex]
T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{\left(\frac{n(n + 1)}{2}\right)}
[/tex]

[tex]
T_n = \frac{1}{2}(n^2 + n)
[/tex]

Hence,

[tex]
S_n = \frac{1}{2}\left(\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\right)
[/tex]

On substituting n = 11, I get:

[tex]
S_n = 286
[/tex]

But this isn't one of the options I got on the test. All the options were between 300-400 and all had their last digit '9' [that's all i remember.. they took the question paper away].
 
Last edited:

Answers and Replies

Gib Z
Homework Helper
3,344
4
Well from what I see there are no errors. Are you sure you have the question right? Even when we sum 12 terms, we get 364, and 13 would be more than 400.
 
412
2
nope.. the question is definately this. I remember it since i tried to solve for it quite some time. And nope.. not even 364 is an option...
 

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