[[1]] is the sum Sn of the cubes of first n terms of an A.P and shows that the sum of the first n terms of the A.P is a factor of Sn .

Question

Infinity Learn · Accepted Answer

Consider, an A.P be (a+d)+(a+2d)+(a+3d)+.........+(a+nd)
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:

Pn = \frac{n}{2} (2 (a + d) + (n - 1) d)

= \frac{n}{2} (2 a + 2 d + nd - d)

= \frac{n}{2} (2 a + nd + d)

= \frac{n}{2} (2 a + (n + 1) d)

Where, Pn is the sum of n terms of the series.
According to the question, Sn is the sum of the cube of first n terms of the series then we get,
Sn=(a+d)3+(a+2d)3+.........+(a+nd)3
By using the identity
(a+b)3=a3+3a2b+3ab2+b3 then we get,
⇒Sn=(a3+3a2d+3ad2+d3)+(a3+6a2d+12ad2+8d3)+.........+(a3+3na2d+3n2ad2+n3d3)
As we know that the series is extend upto nth term but having similar pattern so taking out the common thing then we get,
⇒Sn=na3+3a2d(1+2+3+...+n)+3ad2(12+22+32+....+n2)+d3(13+23+33+....+n3)
Sn=na3+3a2d∑n+3ad2∑n2+d3∑n3
We know that,

\Rightarrow \sum n = \frac{n (n - 1)}{2}

\Rightarrow \sum n 2 = \frac{n (n + 1) (2 n + 1)}{6}

\Rightarrow \sum n 3 = \frac{n^{2} {(n + 1)}^{2}}{4}

Substituting these values in the equation then we get,

\Rightarrow Sn = n a^{3} + \frac{3 a^{2} d^{n} (n - 1)}{2} + \frac{3 a d^{2} n (n + 1) (2 n + 1)}{6} + \frac{d^{3} n^{2} {(n + 1)}^{2}}{4}

This the required equation of sum of cube of first n terms of the A.P series.
Taking

\frac{n}{4}

common from all the terms of the equation then we get,

\Rightarrow Sn = \frac{n}{4} (4 a^{3} + 6 a^{2} d (n - 1) + \frac{a d^{2}}{2} (n + 1) (2 n + 1) + d^{3} n {(n + 1)}^{2})

Taking (2a+(n+1)d) common from the equation then we get,

\Rightarrow Sn = \frac{n}{4} (2 a + (n + 1) d) (2 a^{2} + 2 ad (n + 1) + d^{2} n {(n + 1)}^{2})

Where,

Pn = \frac{n}{2} (2 a + (n - 1) d)

so, substituting the value then we get,

\Rightarrow Sn = \frac{n}{2} Pn (2 a^{2} + 2 ad (n + 1) + d^{2} n {(n + 1)}^{2})

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 the cube of first n terms of the same series.

[[1]] is the sum Sn of the cubes of first n terms of an A.P and shows that the sum of the first n terms of the A.P is a factor of Sn .

Start JEE / NEET / Foundation preparation at rupees 99/day !!

Detailed Solution

tricks from toppers of Infinity Learn

Get Expert Academic Guidance – Connect with a Counselor Today!