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 $A$.$P$. with a common differen

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 $A$.$P$. with a common difference of $d = 2$.First,we find the first term of the $n^{th}$ group. The first terms of the groups are $1, 3, 7, 13, \dots$. Let $a_n$ be the first term of the $n^{th}$ group. The differences between consecutive first terms are $2, 4, 6, \dots$,which form an $A$.$P$.$a_n = a_1 + \sum_{k=1}^{n-1} (2k) = 1 + 2 \cdot \frac{(n-1)n}{2} = 1 + n^2 - n = n^2 - n + 1$.The $n^{th}$ group is an $A$.$P$. with $n$ terms,first term $a = n^2 - n + 1$,and common difference $d = 2$.The sum of the $n^{th}$ group is $S_n = \frac{n}{2} [2a + (n-1)d]$.$S_n = \frac{n}{2} [2(n^2 - n + 1) + (n-1)2]$$S_n = \frac{n}{2} [2n^2 - 2n + 2 + 2n - 2]$$S_n = \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 $5^{th}$ term of a $G.P.$ is $2,$ find the product of first $9$ terms.

If $\sum_{n = 1}^5 \frac{1}{n(n + 1)(n + 2)(n + 3)} = \frac{k}{3}$,then $k$ is equal to

When the $9^{th}$ term of an $A.P.$ is divided by its $2^{nd}$ term,the quotient is $5$. When the $13^{th}$ term is divided by the $6^{th}$ term,the quotient is $2$ and the remainder is $5$. Find the first term of the $A.P.$

Harmonic mean of the roots of the equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ is:

The least positive integer $n$ such that $1 - \frac{2}{3} - \frac{2}{3^2} - \dots - \frac{2}{3^{n-1}} < \frac{1}{100}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module