find the sum of n terms of the series (a+b) + (a2+2b) + (a3 + 3b = .....

The solution to the mentioned question requires a higher level of understanding (concept of G.P.) which is not taught in your grade still i am providing you the answer.

The given series i.e., (a + b) + (a2 + 2b) + (a3 + 3b) + .....+ (an + nb) can be rewritten as

(a + a2 + a3 + ...... + an) + (b + 2b + 3b + ..... + nb)

Clearly the first series forms a G.P. with first term a and common ratio also a.

and the second series forms an A.P. with first term b and common ratio also b.

Hence, required sum is

   for a < 1

or

   for a > 1

  • 40

 a = a+b

d = a2 +2b - a+b = a - b

n = ?

an = ?

an=

  • -19

We have,

(a+b) + (a2+2b) + (a3 + 3b)...to n terms

= (a + a2 + a3 + ... to n terms) + b (1 + 2 + 3 +...to n terms)

= a(1- a 1)/ 1- a    + b  {n(n+1)/ 2}


  • 12

cant understand

  • -12
What are you looking for?