The sum of (n - 1) terms of the series 1 + (1 | vedclass

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

Question

(C) Let $T_k$ be the $k^{th}$ term of the series.The $k^{th}$ term is the sum of the first $k$ odd numbers: $T_k = 1 + 3 + 5 + \dots + (2k - 1) = k^2$

Vedclass · Accepted Answer

$\frac{n(n - 1)(2n - 1)}{6}$

Vedclass · Answer

(C) Let $T_k$ be the $k^{th}$ term of the series.The $k^{th}$ term is the sum of the first $k$ odd numbers: $T_k = 1 + 3 + 5 + \dots + (2k - 1) = k^2$.We need to find the sum of $(n - 1)$ terms,which is $S_{n-1} = \sum_{k=1}^{n-1} T_k = \sum_{k=1}^{n-1} k^2$.Using the formula for the sum of squares of the first $m$ natural numbers,$\sum_{k=1}^{m} k^2 = \frac{m(m + 1)(2m + 1)}{6}$.Substituting $m = n - 1$:$S_{n-1} = \frac{(n - 1)((n - 1) + 1)(2(n - 1) + 1)}{6} = \frac{(n - 1)(n)(2n - 2 + 1)}{6} = \frac{n(n - 1)(2n - 1)}{6}$.

The sum of $(n - 1)$ terms of the series $1 + (1 + 3) + (1 + 3 + 5) + \dots$ is

Similar Questions

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 $A.P.$

If the first term of a $G.P.$ is $5$ and the common ratio is $-5$,then which term is $3125$ (in $^{th}$)?

Let $f: R \rightarrow R$ be such that for all $x \in R$,$(2^{1+x}+2^{1-x})$,$f(x)$,and $(3^x+3^{-x})$ are in $A.P.$. Then,the minimum value of $f(x)$ is:

If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$, then the greatest number of the series will be:

If the sum of three numbers of an arithmetic sequence is $15$ and the sum of their squares is $83$,then the numbers are

Vedclass Test Series

Exam Paper Generator

Online Exam Module