The sum of the first 9 terms of the series fr | vedclass

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

Question

(C) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n (2k-1)}$.We know that $\sum_{k=1}^n k^3 = \left[\frac{n(n+

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(C) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n (2k-1)}$.We know that $\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2$ and the sum of the first $n$ odd numbers is $\sum_{k=1}^n (2k-1) = n^2$.Thus,$T_n = \frac{\left[\frac{n(n+1)}{2}\right]^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}$.We need to find the sum of the first $9$ terms,$S_9 = \sum_{n=1}^9 T_n = \sum_{n=1}^9 \frac{(n+1)^2}{4}$.$S_9 = \frac{1}{4} \sum_{n=1}^9 (n+1)^2 = \frac{1}{4} (2^2 + 3^2 + \dots + 10^2)$.Adding and subtracting $1^2$,we get $S_9 = \frac{1}{4} \left[ \sum_{k=1}^{10} k^2 - 1^2 \right]$.Using the formula $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$,for $m=10$:$S_9 = \frac{1}{4} \left[ \frac{10(11)(21)}{6} - 1 \right] = \frac{1}{4} [385 - 1] = \frac{384}{4} = 96$.

The sum of the first $9$ terms of the series $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ is:

Similar Questions

If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots = \frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in R$,then $(a-b)^2=$

The $9^{th}$ term of the sequence $27, 9, 5\frac{2}{5}, 3\frac{6}{7}, \dots$ is $.....$

The sequence of natural numbers is divided into groups as follows: $(1), (2, 3), (4, 5, 6), (7, 8, 9, 10), \dots$. Find the sum of the numbers in the $n^{th}$ group.

Find the fractional value of $0.1232323......$

The numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module