Question

Find the sum of the following geometric series :

(i) (x + y) + (x² + xy + y²) + (x³ + x² y + xy² + y³) + …. to n terms;

(ii) 3/5 + 4/5² + 3/5³ + 4/5⁴ + … to 2n terms;

Answer 1

(i) (x + y) + (x² + xy + y²) + (x³ + x² y + xy² + y³) + …. to n terms;

Let S_n = (x + y) + (x² + xy + y²) + (x³ + x² y + xy² + y³) + …. to n terms

Let us multiply and divide by (x – y) we get,

S_n = 1/(x – y) [(x + y) (x – y) + (x² + xy + y²) (x – y) … upto n terms]

(x – y) S_n = (x² – y²) + x³ + x²y + xy² – x²y – xy² – y³..upto n terms

(x – y) S_{n =} (x² + x³ + x⁴+…n terms) – (y² + y³ + y⁴ +…n terms)

By using the formula,

Sum of GP for n terms = a(1 – rⁿ )/(1 – r)

We have two G.Ps in above sum, so,

(x – y) S_n = x² [(xⁿ – 1)/ (x – 1)] – y² [(yⁿ – 1)/ (y – 1)]

S_n = 1/(x - y) {x² [(xⁿ – 1)/ (x – 1)] – y² [(yⁿ – 1)/ (y – 1)]}

(ii) 3/5 + 4/5² + 3/5³ + 4/5⁴ + … to 2n terms;

The series can be written as:

3 (1/5 + 1/5³ + 1/5⁵+ … to n terms) + 4 (1/5² + 1/5⁴ + 1/5⁶ + … to n terms)

Firstly let us consider 3 (1/5 + 1/5³ + 1/5⁵+ … to n terms)

So, a = 1/5

r = t₂/t₁ = 1/5² = 1/25

By using the formula,

Sum of GP for n terms = a(1 – rⁿ )/(1 – r)

Now, Let us consider 4 (1/5² + 1/5⁴ + 1/5⁶ + … to n terms)

So, a = 1/25

r = t₂/t₁ = 1/5² = 1/25

By using the formula,

Sum of GP for n terms = a(1 – rⁿ )/(1 – r)