For any odd integer n ge 1,{n^3} - {(n - 1)^3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $S = {n^3} - {(n - 1)^3} + {(n - 2)^3} - {(n - 3)^3} + \dots + {1^3}$.Since $n$ is odd,the last term is ${1^3}$.We can write $S = \sum_{k=1}^{

Vedclass · Accepted Answer

$\frac{1}{4}{(n + 1)^2}(2n - 1)$

Vedclass · Answer

(D) Let $S = {n^3} - {(n - 1)^3} + {(n - 2)^3} - {(n - 3)^3} + \dots + {1^3}$.Since $n$ is odd,the last term is ${1^3}$.We can write $S = \sum_{k=1}^{n} k^3 - 2 \sum_{k=1}^{(n-1)/2} (2k)^3$.Using the formula $\sum_{k=1}^{m} k^3 = \left[ \frac{m(m+1)}{2} ight]^2$,we get:$S = \left[ \frac{n(n+1)}{2} ight]^2 - 2 	imes 8 \sum_{k=1}^{(n-1)/2} k^3$$S = \frac{n^2(n+1)^2}{4} - 16 \left[ \frac{\frac{n-1}{2} (\frac{n-1}{2} + 1)}{2} ight]^2$$S = \frac{n^2(n+1)^2}{4} - 16 \left[ \frac{(n-1)(n+1)}{8} ight]^2$$S = \frac{n^2(n+1)^2}{4} - 16 \frac{(n-1)^2(n+1)^2}{64}$$S = \frac{n^2(n+1)^2}{4} - \frac{(n-1)^2(n+1)^2}{4}$$S = \frac{(n+1)^2}{4} [n^2 - (n-1)^2]$$S = \frac{(n+1)^2}{4} [n^2 - (n^2 - 2n + 1)]$$S = \frac{(n+1)^2}{4} (2n - 1)$.

For any odd integer $n \ge 1$,${n^3} - {(n - 1)^3} + \dots + {( - 1)^{n - 1}}{1^3} = $

Similar Questions

If $S$ is the sum of $n$ terms of the series $1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$,then $S = \dots$

What is the sum of the series $1 + (1 + x) + (1 + x + x^2) + (1 + x + x^2 + x^3) + \dots$ up to $n$ terms?

Find the sum of the first $n$ terms of the series: $3+7+13+21+31+\ldots$

If $1 + \sin \theta + \sin^2 \theta + \dots \text{ to } \infty = 4 + 2\sqrt{3}$,where $0 < \theta < \pi$ and $\theta \neq \frac{\pi}{2}$,then $\theta = $

The minimum value of $n$ for which $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2} < 1.01$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module