given that 3x-5y=7, what is (6-3x+5y)+(6-3x+5y)^2+..........+(6-3x+5y)^2002?
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calendarEducator since 2010
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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.
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
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.
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