If sum_{r=1}^{n}r^3 - sum_{p=1}^{n}sum_{m=1}^ | vedclass

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

Question

(B) We are given the equation $\sum_{r=1}^{n}r^3 - \sum_{p=1}^{n}\sum_{m=1}^{p}\sum_{r=1}^{m}1 = 80$. First,we know that $\sum_{r=1}^{n}r^3 = \left[ \

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) We are given the equation $\sum_{r=1}^{n}r^3 - \sum_{p=1}^{n}\sum_{m=1}^{p}\sum_{r=1}^{m}1 = 80$. First,we know that $\sum_{r=1}^{n}r^3 = \left[ \frac{n(n+1)}{2} \right]^2$. Next,consider the triple summation: $\sum_{p=1}^{n}\sum_{m=1}^{p}\sum_{r=1}^{m}1 = \sum_{p=1}^{n}\sum_{m=1}^{p} m = \sum_{p=1}^{n} \frac{p(p+1)}{2} = \frac{1}{2} \left[ \sum_{p=1}^{n} p^2 + \sum_{p=1}^{n} p \right]$. Using standard summation formulas: $\frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] = \frac{n(n+1)}{4} \left[ \frac{2n+1}{3} + 1 \right] = \frac{n(n+1)}{4} \left[ \frac{2n+4}{3} \right] = \frac{n(n+1)(n+2)}{6} = \binom{n+2}{3}$. Substituting back into the original equation: $\left[ \frac{n(n+1)}{2} \right]^2 - \frac{n(n+1)(n+2)}{6} = 80$. For $n=4$: $\left[ \frac{4(5)}{2} \right]^2 - \frac{4(5)(6)}{6} = 10^2 - 20 = 100 - 20 = 80$. Thus,$n=4$ is the correct value.

If $\sum_{r=1}^{n}r^3 - \sum_{p=1}^{n}\sum_{m=1}^{p}\sum_{r=1}^{m}1 = 80$,then the possible value of $n$ is:

Similar Questions

The sum of the series $3 + 33 + 333 + \dots$ to $n$ terms is

The value of $1^3 - 2^3 + 3^3 - \dots + 15^3$ is:

Find the sum of the following series up to $n$ terms:
$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$

If the sum of the first $n$ terms of a series is $5n^2 + 2n$,then its second term is

Vedclass Test Series

Exam Paper Generator

Online Exam Module