Question

If Sn=1+3+6+10+...+n(n+1)/2 then Sn is

Answer 1

Hey friend, Harish here.

Here is your answer.

$\mathrm{S_n = 1+3+6+...\frac{n(n+1)}{2}} \\ \\ \sum\limits_{k=1}^n\Bigl( \frac{k(k+1)}{2} \Bigr) = \frac{1}{2} \sum\limits_{k=1}^n\Bigl( k^2 + k \Bigr ) \\ \\ \\ \implies \frac{1}{2} \sum\limits_{k=1}^n\Bigl( k^2 \Bigr ) + \sum\limits_{k=1}^n\Bigl( k \Bigr) \\ \\ \\ \implies \frac{1}{2} \Bigl( \frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2} \Bigr) \\ \\ \\ \implies \frac{1}{2} \Bigl( \frac{n\left(n+1\right)\left(n+2\right)}{3} \Bigr) \\ \\ \\ \boxed{\bold{\frac{n(n+1)(n+2)}{6}}}$

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Hope my answer is helpful to you.

Answer 2

Answer:

The sum of n term is $S_n=\frac{n(n+1)(n+2)}{6}}$    

Step-by-step explanation:

Given : $S_n=1+3+6+10+.....\frac{n(n+1)}{2}$

To find : The value of $S_n$?

Solution :

$S_n=1+3+6+10+.....\frac{n(n+1)}{2}$

We have given the nth term of the sequence,

$T_n=\frac{n(n+1)}{2}$

Sum of n terms is given by,

$S_n=\sum\limits_{k=1}^n(T_n)$

$S_n=\sum\limits_{k=1}^n\Bigl( \frac{k(k+1)}{2} \Bigr)$

$S_n=\sum\limits_{k=1}^n\Bigl(\frac{k^2 + k}{2} \Bigr )$

$S_n=\sum\limits_{k=1}^n\Bigl( k^2 + k \Bigr )$

$S_n=\frac{1}{2}\sum\limits_{k=1}^n\Bigl( k^2 \Bigr )+\sum\limits_{k=1}^n\Bigl( k \Bigr)$

$S_n=\frac{1}{2}\Bigl( \frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\Bigr)$

$S_n=\frac{1}{2}\Bigl( \frac{n\left(n+1\right)\left(n+2\right)}{3} \Bigr)$

$S_n=\frac{n(n+1)(n+2)}{6}}$

Therefore, The sum of n term is $S_n=\frac{n(n+1)(n+2)}{6}}$