Question

Find the sum Sn of the first n terms from the given sequence.
1, 1+2, 1+2+3, 1+2+3+4, ...
Please show your work and use Sigma Notation too. Thank you!​

Answer 1

The Sequence:

1 , 1+2 , 1+2+3 , 1+2+3+4, ...

here, every term is an AP. finding the general formula for the sum of the elements of each term is:

Sₙ = $\frac{x}{2}(2a + (x-1)d)$        [where x = number of term, a = first term and d = common difference]

here, the first term is always 1 and so is the common difference.

Sₙ = $\frac{x}{2}(2 +x-1)$

Sₙ = $\frac{x}{2}(1 +x)$ = $\frac{1}{2}(x + x^{2})$

which is the formula for a general term in our series

now, we need to find the sum of the first n terms of this series

$\displaystyle\sum_{x=1}^{n} [\frac{1}{2}(x + x^{2})]$

$\displaystyle\frac{1}{2} [\sum_{x=1}^{n} (x) + \sum_{x=1}^{n}(x^{2})]$

in this formula, for the first term, it's just an AP from x = 1 to x = n

for the second term, we have a general formula $\frac{n(n+1)(2n+1)}{6}$

$\frac{1}{2}[\frac{n}{2}(2a + (n-1)d)+ \frac{n(n+1)(2n+1)}{6} ]$

in this AP (first term), the first term and the common difference is 1 as well

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

$\frac{1}{2}[\frac{n}{2}(n+1)+ \frac{n(n+1)(2n+1)}{6} ]$

$[\frac{n}{4}(n+1)+ \frac{n(n+1)(2n+1)}{12} ]$

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

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

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

$\frac{n}{2}(n+1) [\frac{(n+2)}{3} ]$

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

which is the sum of n terms of the given sequence