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 =
`(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 =
`(x)/(2)(2 +x-1)`

Sₙ =
`(x)/(2)(1 +x)` =
`(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) [(1)/(2)(x + x^(2))]`

`\displaystyle(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
`(n(n+1)(2n+1))/(6)`

`(1)/(2)[(n)/(2)(2a + (n-1)d)+ (n(n+1)(2n+1))/(6) ]`

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

`(1)/(2)[(n)/(2)(2 + n-1)+ (n(n+1)(2n+1))/(6) ]`

`(1)/(2)[(n)/(2)(n+1)+ (n(n+1)(2n+1))/(6) ]`

`[(n)/(4)(n+1)+ (n(n+1)(2n+1))/(12) ]`

`(n)/(4)(n+1) [1+((2n+1))/(3) ]`

`(n)/(4)(n+1) [((3+2n+1))/(3) ]`

`(n)/(4)(n+1) [((2n+4))/(3) ]`

`(n)/(2)(n+1) [((n+2))/(3) ]`

`(n(n+1)(n+2))/(6)`

which is the sum of n terms of the given sequence