# given that 3x-5y=7, what is (6-3x+5y)+(6-3x+5y)^2+..........+(6-3x+5y)^2002?

*print*Print*list*Cite

We have to find the sum of (6-3x+5y)+(6-3x+5y)^2+..........+(6-3x+5y)^2002, given that 3x - 5y = 7.

Now we see that the terms of the sum are in a GP. The first term is 6 - 3x + 5y and each subsequent term is a product of the previous with 6 - 3x + 5y

As 3x - 5y = 7

=> 6 - 3x + 5y

=> -1

So the required sum is -1 + 1 - 1 + 1...-1^2002.

=> a*(r^(n+1) - 1)/ (r - 1)

=> -1*((-1)^ 2002 - 1)/(-1-1)

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

=> 0

**Therefore (6-3x+5y)+(6-3x+5y)^2+..........+(6-3x+5y)^2002 = 0.**

3x-5y = 7

We need to find the value of the expression:

E = (6-3x+5y) + (6-3x_5y)^2 + ....+ (6-3x+5y)^2002

We will rewrite:

E = ( 6-(3x-5y)) + (6-(3x-5y)^2 + ....+(6-(3x-5y))^2002

Now we know that 3x-5y = 7

==> E = (6-7) + (6-7)^2 + ....+(6-7)^2002

==> E = -1 + (-1)^2 + (-1)^3 + ....+ (-1)^2001+ (-1)^2002

==> E = -1 + 1 -1 + 1 .... + -1 + 1 = 0

**Then the values of the expression (6-3x+5y)+(6-3x+5y)^2+ ...(6-3x+5y)^2002 = 0**

3x-5y = 7.

Therefore 6-3x+5y).

Therefore 6-(3x-5y = 6-7 =-1.

Therefore (-3x+5y) = -1. We substitute 6-3x+5) = -1 in (6-3x+5y)+ (6-3x+5y)^2+ (6-3x+5y)^2+...... +(6-3x+5y)^2002 and get:

(-1)+(-1)^2+(-1)^3+(-1)^4+....+(-1)^2002. There are 2002 terms.

(-1+1)+(-1+1)+(-1+1).....(-1+1) = 0, as there are 1001 pairs of brackets In each brackes.

Therefore (6-3x+5y)+(6-3x+5y)^2+..........+(6-3x+5y)^2002 = 0.