Find the sum of geometric series (x + y) + (x2 + xy + y2) + (x3 + x2 y + xy2 + y3) + …. to n terms

Asked by Sakshi | 1 year ago |  80

##### Solution :-

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

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

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

Sn = $$\dfrac{ 1}{(x – y) [(x + y) (x – y) + (x^2 + xy + y^2) (x – y)}$$ … upto n terms]

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

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

By using the formula,

Sum of GP for n terms = $$\dfrac{ a(1 – r^n )}{(1 – r)}$$

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

Sn = $$\dfrac{1}{(x-y) {x^2 [(x^n – 1)}{ (x – 1)] – y^2 [(y^n – 1)} (y – 1)]}$$

Answered by Sakshi | 1 year ago

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