simplify (πd²/4)-(π[d+1]²/4)

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simplify (πd²/4)-(π[d+1]²/4)

Mathematics
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Factor pi/4. Then, it is a difference of squares. Simplify.
wouldn't it be π/4 (d^2-d^2+1)?
No. It is pi/4(d^2 - (d+1)^2)

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so how do i solve that? :S
a^2 - b^2 = (a-b)(a+b)
In the above, a = d, b = d+1.
i got π(2d-1)/4 .... is that right? because the back of my book says the answer is π(2d+1)/4
pi/4 (d^2 - (d+1)^2) = pi/4 (d + d + 1)(d - (d-1)) = pi/4 (2d+1)(d-d+1) = pi/4 (2d+1)
how is it pi/4 (d + d + 1)(d - (d-1)) ?
Forget pi/4 since that is already factored out. Focus on (d^2 - (d+1)^2). Like I said, a^2 - b^2 = (a+b)(a-b). Here, you have d^2 - (d+1)^2 = (d+d+1)(d-(d+1))
OOOOOH ! im subtracting it.. instead of multiplying (x my bad
wait but i still get 2d-1 :S
Actually, you get -(2d+1)
): it says the answer is pi/4 2d+1
I see the answer as the same, but with a minus sign. But, may be missing something.
oh well i'll just keep it like this for now and ask my teacher tomorrow. thanks

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