# What is the result of the sum (5-2a+3b)+(5-2a+3b)^2+...+(5-2a+3b)^500 if 2a-3b=6 ?

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We notice that the terms of the sum are the terms of geometric progression. The first term is 5-2a+3b and each consecutive term is a product between the previous term, multiplied by 5-2a+3b.

Since 2a - 3b = 6 => 5-2a+3b = 5 - 6 = -1

So the requested sum is -1 + 1 - 1 + 1...(-1)^500.

We'll use the formula of the sum of the first 500 terms of a g.p.

Sn = b*(q^n - 1)/ (q - 1)

S500 = -1*[(-1)^ 500 - 1]/(-1-1)

S500 = -1*(1-1)/-2

S500 = 0

**Therefore, the result of the sum is: (5-2a+3b)+(5-2a+3b)^2+...+(5-2a+3b)^500 = 0.**

The value of 2a - 3b = 6

We expression we have to find the sum of is (5- 2a + 3b)+(5- 2a + 3b)^2+...+(5 - 2a + 3b)^500

(5- 2a + 3b)+(5- 2a + 3b)^2+...+(5 - 2a + 3b)^500

=> (5- (2a - 3b))+(5- (2a - 3b))^2+...+(5 - (2a - 3b))^500

=> (5- 6)+(5- 6)^2+...+(5 - 6)^500

=> -1 + -1^2 + -1^3 +...+ -1^500

=> -1 + 1 - 1 + 1 - 1 +... + 1

The exponents of -1 are from 1 to 500. From 1 to 500 we an equal number of even and odd numbers.

**Therefore the sum of the terms is equal to 0. We get (5- (2a - 3b))+(5- (2a - 3b))^2+...+(5 - (2a - 3b))^500 = 0**