The sum of all terms of the n^{th} bracket of | vedclass

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

Question

(D) The number of terms in successive groups are $1, 2, 3, \dots, n$. Thus,the $n^{th}$ group contains $n$ terms in an arithmetic progression with a c

Vedclass · Accepted Answer

$n^3$

Vedclass · Answer

(D) The number of terms in successive groups are $1, 2, 3, \dots, n$. Thus,the $n^{th}$ group contains $n$ terms in an arithmetic progression with a common difference $d = 2$.First,we find the first term of the $n^{th}$ group. Let $a_n$ be the first term of the $n^{th}$ group. The sequence of first terms is $1, 3, 7, 13, \dots$. The differences between consecutive terms are $2, 4, 6, \dots$,which form an arithmetic progression.Let $a_n = a_1 + \sum_{k=1}^{n-1} (2k) = 1 + 2 \cdot \frac{(n-1)n}{2} = 1 + n^2 - n = n^2 - n + 1$.The $n^{th}$ group is an arithmetic progression with $n$ terms,first term $a = n^2 - n + 1$,and common difference $d = 2$.The sum of the $n^{th}$ group is $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values: $S_n = \frac{n}{2} [2(n^2 - n + 1) + (n - 1)2] = \frac{n}{2} [2n^2 - 2n + 2 + 2n - 2] = \frac{n}{2} [2n^2] = n^3$.

The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:

Similar Questions

The sum $\frac{3 \times 1}{1^2} + \frac{5 \times (1^3 + 2^3)}{1^2 + 2^2} + \frac{7 \times (1^3 + 2^3 + 3^3)}{1^2 + 2^2 + 3^2} + \dots$ up to the $10^{th}$ term is:

$2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdot 16^{1/32} \cdots$ is equal to

For all $n \in N$,if $1^3+2^3+3^3+\ldots+n^3 > x$,then a value of $x$ among the following is

What is the sum of the first $10$ terms of the series $0.7 + 0.77 + 0.777 + \dots$?

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module