Question

What is the sum upto n terms of the series 3c1 +7c2 +11c3..?

Answer 1

Answer:

S = [-c + 4c(1 - cⁿ)/(1 - c) -  [3 + 4(n-1)]cⁿ⁺¹]/(1-c)

Step-by-step explanation:

Hi.

The series of the above form is known as Arithmetic-Geometric Progression, where part of the trms would be in A.P and the rest of the terms would form a G.P.

From the above series, we can clearly observe that the coefficient of

powers of c in each term form an A.P , hence its nth term coefficient would

be 3 + 4(n-1).

Whenever, a question is given on A.G.P, it is advisable to follow the below procedure to solve it,

Let     S =   3c + 7c² + 11c³ + .......+[3 + 4(n-1)]cⁿ,

then cS =           3c² + 7c³ + ...............................+[3 + 4(n-1)]cⁿ⁺¹

Subtracting the second one from first, we get,

(1 - c)S = 3c + 4c² + 4c³ +..........+ 4cⁿ - [3 + 4(n-1)]cⁿ⁺¹

(1 - c)S = -c + 4c + 4c² + 4c³ +..........+ 4cⁿ - [3 + 4(n-1)]cⁿ⁺¹

starting from 2nd term the next n terms form a G.P, hence using sum to nterms of G.P formula, we get

(1 - c)S = -c + 4c(1 - cⁿ)/(1 - c) -  [3 + 4(n-1)]cⁿ⁺¹

=> S = [-c + 4c(1 - cⁿ)/(1 - c) -  [3 + 4(n-1)]cⁿ⁺¹]/(1-c)

Hope, it helped !

Answer 2

Answer:

2^n(2^n - 1) + 1

Step-by-step explanation: