Question

Find the sum of n terms of the series (a+b) + ( a2 + 2b) + (a3+3b)+

Answer 1

(Considering the terms as - a² + 2b)

On opening the brackets,
The series forms a GP in a, with common ratio 'a' and first term 'a'.
a + a² + a³ + a⁴ +...
And
The series forms an AP in b, with common difference 'b' and first term b.
b + 2b + 3b + 4b +...

Last term of the series will be (a^n + nb)


Sum of n terms of AP= (n/2)(first term + last term)
Sum of n terms of GP= [ first term * (ratio^n -1) ] / (ratio - 1)

Thus, on substituting the values in these formulae,
We get the sum of the sequence as =
[a*(a^n -1)]/(a-1) + [b*n*(n+1)]/2

Answer 2

a(1+2+...n) + b(1+2+3+..n)
(a+b)(n(n-1)/2)
Using ap