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## 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

## 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]

## 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].

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