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

If $\sum_{r=1}^{10} r! (r^3 + 6r^2 + 2r + 5) = \alpha(11!)$,then the value of $\alpha$ is equal to ...... .

Consider an infinite $G.P.$ with first term $a$ and common ratio $r$. Its sum is $4$ and the second term is $3/4$. Then:

$1 + (1 + a)x + (1 + a + a^2)x^2 + \dots \infty = \dots \, (0 < a, x < 1)$

Find the sum to $n$ terms of the series $3 \times 8 + 6 \times 11 + 9 \times 14 + \dots$

If the $n^{th}$ term of a sequence is $n(n + 1)$,then what is the sum of its $n$ terms?

Vedclass Test Series

Exam Paper Generator

Online Exam Module