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 sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is

If $3 + 3\alpha + 3\alpha^2 + \dots \infty = \frac{45}{8}$,then the value of $\alpha$ will be

$1 + (1 + a)x + (1 + a + a^2)x^2 + \dots \infty = \dots \, (0 < a, x < 1)$

The sum of $n$ terms of the following series $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \dots$ is:

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module