Question

Find the sum of the cubes of an A.P., and show that it is exactly divisible by the sum of the terms.

Answer 1

Here the A.P. is not given with the specific terms. so we can take the sum of the $n$ terms. but the restriction is that its cube is to be found. Thus we will generalize the formula and then find whether the sum is divisible or not.

Complete step by step answer:
We can write the sum of cubes of an A.P. as,
$S = \sum {{{(a + (n - 1)d)}^3}}$
Now as we know that the cubic equation can be written as, ${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$.
And in general, on expanding we get,
$S = n{a^3} + \dfrac{{3{a^2}d\left( {n - 1} \right)n}}{2} + \dfrac{{3a{d^2}\left( {n - 1} \right)n\left( {2n - 1} \right)}}{6} + \dfrac{{{d^3}{{\left( {n - 1} \right)}^2}{n^2}}}{4}$
Taking $\dfrac{n}{4}$ common,
$S = \dfrac{n}{4}\left[ {4{a^3} + 6{a^2}d\left( {n - 1} \right) + 2a{d^2}\left( {n - 1} \right)\left( {2n - 1} \right) + {d^3}n{{\left( {n - 1} \right)}^2}} \right]$
On rearranging the terms as per to get the formula for sum of n terms we get.
$S = \dfrac{n}{4}\left[ {2a + \left( {n - 1} \right)d} \right]\left[ {2{a^2} + 2\left( {n - 1} \right)ad + n\left( {n - 1} \right){d^2}} \right]$
Now taking 2 common from the last bracket ,
$S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\left[ {{a^2} + \left( {n - 1} \right)ad + \dfrac{{n\left( {n - 1} \right){d^2}}}{2}} \right]$
As we know that n and n-1 are two consecutive terms in an A.P. or in general then it is absolutely divisible by 2.

Thus the sum is exactly divisible by $\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.

Note: we have to take the sum of cubes of an A.P. and not the cubes of the first three terms. This can be confusing for students whether to take the sum of all cubes or cubes of particular terms. so carefully read the question. We know that $a + (n - 1)d$ is the formula for the last term of an A.P. we used that because no particular number of terms is given.We know that,

\Rightarrow \sum\limits_{k = 1}^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \\\ \Rightarrow \sum\limits_{k = 1}^n {{k^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} $$ Can be used if required.