Question

Find the sum $ {S_n} $ of the cubes of first n terms of an A.P and show that the sum of the first n terms of the A.P is a factor of $ {S_n} $ .

Answer 1

Hint: The sum of any n terms of the arithmetic progression series is carried out by using the formula $ {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) $ where, n is the number of terms in series, a and d are for first term and common difference.

Complete step-by-step answer:
Consider, an A.P be $ \left( {a + d} \right) + \left( {a + 2d} \right) + \left( {a + 3d} \right) + ......... + \left( {a + nd} \right) $
Here, a common difference is not added with the first terms.
As we know that he sum of the A.P is carried out as:
  $
{P_n} = \dfrac{n}{2}\left( {2\left( {a + d} \right) + \left( {n - 1} \right)d} \right)\\
 = \dfrac{n}{2}\left( {2a + 2d + nd - d} \right)\\
 = \dfrac{n}{2}\left( {2a + nd + d} \right)\\
 = \dfrac{n}{2}\left( {2a + \left( {n + 1} \right)d} \right)
  $
Where, $ {P_n} $ is the sum of n terms of the series.
According to the question, $ {S_n} $ is the sum of the cube of first n terms of the series then we get,
  $ {S_n} = {\left( {a + d} \right)^3} + {\left( {a + 2d} \right)^3} + ......... + {\left( {a + nd} \right)^3} $
By using the identity
 $ {\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} $ then we get,
$\Rightarrow {S_n} = \left( {{a^3} + 3{a^2}d + 3a{d^2} + {d^3}} \right) + \left( {{a^3} + 6{a^2}d + 12a{d^2} + 8{d^3}} \right) + ......... + \left( {{a^3} + 3n{a^2}d + 3{n^2}a{d^2} + {n^3}{d^3}} \right)\\
\Rightarrow {S_n} = {a^3} + 3{a^2}d + 3a{d^2} + {d^3} + {a^3} + 6{a^2}d + 12a{d^2} + 8{d^3} + ......... + {a^3} + 3n{a^2}d + 3{n^2}a{d^2} + {n^3}{d^3}$
As we know that the series is extend upto nth term but having similar pattern so taking out the common thing then we get,
 $ \Rightarrow {S_n} = n{a^3} + 3{a^2}d\left( {1 + 2 + 3 + ... + n} \right) + 3a{d^2}\left( {{1^2} + {2^2} + {3^2} + .... + {n^2}} \right) + {d^3}\left( {{1^3} + {2^3} + {3^3} + .... + {n^3}} \right)\\
{S_n} = n{a^3} + 3{a^2}d\sum n + 3a{d^2}\sum {{n^2}} + {d^3}\sum {{n^3}} $
We know that,
  $
\Rightarrow \sum n = \dfrac{{n\left( {n - 1} \right)}}{2}\\
\Rightarrow \sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\
\Rightarrow \sum {{n^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}
  $
Substituting these values in the equation then we get,
$\Rightarrow {S_n} = n{a^3} + 3{a^2}d\dfrac{{n\left( {n - 1} \right)}}{2} + 3a{d^2}\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + {d^3}\dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} $
This the required equation of sum of cube of first n terms of the A.P series.
Taking $ \dfrac{n}{4} $ common from all the terms of the equation then we get,
$\Rightarrow {S_n} = \dfrac{n}{4}\left( {4{a^3} + 6{a^2}d\left( {n - 1} \right) + \dfrac{{a{d^2}}}{2}\left( {n + 1} \right)\left( {2n + 1} \right) + {d^3}n{{\left( {n + 1} \right)}^2}} \right) $
Taking $ \left( {2a + \left( {n + 1} \right)d} \right) $ common from the equation then we get,
$\Rightarrow {S_n} = \dfrac{n}{4}\left( {2a + \left( {n + 1} \right)d} \right)\left( {2{a^2} + 2ad\left( {n + 1} \right) + {d^2}n{{\left( {n + 1} \right)}^2}} \right) $
Where, $ {P_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) $ so, substituting the value then we get,
 $\Rightarrow {S_n} = \dfrac{n}{2}{P_n}\left( {2{a^2} + 2ad\left( {n + 1} \right) + {d^2}n{{\left( {n + 1} \right)}^2}} \right) $
Hence we can say that the sum of the n terms of the series which is in A.P form is the factor of the sum of cube of first n terms of the same series.

Note: The series for cube and square of first n terms of special series or arithmetic series can be carried out directly by using the summation formulas which are as follow:
  $
\sum n = \dfrac{{n\left( {n - 1} \right)}}{2}\\
\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\
\sum {{n^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}
  $