Find the sum to n terms of the series 1^{2} + | vedclass

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

Question

(A) The $k$-th term of the series is $a_{k} = \sum_{i=1}^{k} i^{2} = \frac{k(k+1)(2k+1)}{6}$.Expanding this,we get $a_{k} = \frac{2k^{3} + 3k^{2} + k}

Vedclass · Accepted Answer

$\frac{n(n+1)^{2}(n+2)}{12}$

Vedclass · Answer

(A) The $k$-th term of the series is $a_{k} = \sum_{i=1}^{k} i^{2} = \frac{k(k+1)(2k+1)}{6}$.Expanding this,we get $a_{k} = \frac{2k^{3} + 3k^{2} + k}{6} = \frac{1}{3}k^{3} + \frac{1}{2}k^{2} + \frac{1}{6}k$.The sum to $n$ terms is $S_{n} = \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} (\frac{1}{3}k^{3} + \frac{1}{2}k^{2} + \frac{1}{6}k)$.Using the standard summation formulas $\sum k^{3} = \frac{n^{2}(n+1)^{2}}{4}$,$\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$,and $\sum k = \frac{n(n+1)}{2}$:$S_{n} = \frac{1}{3} \cdot \frac{n^{2}(n+1)^{2}}{4} + \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{1}{6} \cdot \frac{n(n+1)}{2}$.Factor out $\frac{n(n+1)}{12}$:$S_{n} = \frac{n(n+1)}{12} [n(n+1) + (2n+1) + 1] = \frac{n(n+1)}{12} [n^{2} + n + 2n + 2] = \frac{n(n+1)}{12} [n(n+1) + 2(n+1)]$.$S_{n} = \frac{n(n+1)(n+1)(n+2)}{12} = \frac{n(n+1)^{2}(n+2)}{12}$.

Find the sum to $n$ terms of the series $1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + \ldots$

Similar Questions

The ${n^{th}}$ term 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:

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:

If $\operatorname{gcd}(m, n) = 1$ and $1^2 - 2^2 + 3^2 - 4^2 + \ldots + (2021)^2 - (2022)^2 + (2023)^2 = 1012 m^2 n$,then $m^2 - n^2$ is equal to

In the sequence of sets $(1,2,3), (4,5,6), (7,8,9,10), \ldots$,the sum of elements in the $50^{th}$ set is

Let $A = \sum_{i=1}^{10} \sum_{j=1}^{10} \min \{i, j\}$ and $B = \sum_{i=1}^{10} \sum_{j=1}^{10} \max \{i, j\}$. Then $A + B$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module