Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

2. Find the sum to n terms of each of the series in $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + ...$

Answers (1)

best_answer

the series = $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + ...$

n th term = $n(n+1)(n+2)=a_n$

$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k(k+1)(k+2)$

$=\sum _{k=1}^{n} k^3+3\sum _{k=1}^{n} k^2+2\sum _{k=1}^{n} k$

$=\left [ \frac{n(n+1)}{2} \right ]^2+\frac{3.n(n+1)(2n+1)}{6}+\frac{2.n(n+1)}{2}$

$=\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{2}+n(n+1)$

$=\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+(2n+1)+2)$

$=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{n^2+n+4n+2+4}{2} \right )$

$=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{n^2+5n+6}{2} \right )$

$=\left [ \frac{n(n+1)}{4} \right ] \left ( n^2+5n+6 \right )$

$=\left [ \frac{n(n+1)}{4} \right ] \left ( n^2+2n+3n+6 \right )$

$=\left [ \frac{n(n+1)}{4} \right ] \left ( n(n+2)+3(n+2)\right )$

$=\left [ \frac{n(n+1)}{4} \right ] \left ( (n+2)(n+3)\right )$

$=\left [ \frac{n(n+1)(n+2)(n+3)}{4} \right ]$

Thus, sum is

$=\left [ \frac{n(n+1)(n+2)(n+3)}{4} \right ]$

Posted by

seema garhwal

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Latest Question