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Eva8 [605]
3 years ago
9

Using suitable algebraic identities, find the value of 2004^2 - 2003^2 + 2002^2 - 2001^2

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
3 0

(x+1)^2-x^2=(x^2+2x+1)-x^2=2x+1

Let x=2003. Then

2004^2-2003^2=(2003+1)^2-2003^2=2\cdot2003+1=4007

Similarly, if x=2001, then

2002^2-2001^2=2\cdot2001+1=4003

So

2004^2-2003^2+2002^2-2001^2=4007+4003=8010

Alternatively, you can consider the larger expression

(x+3)^2-(x+2)^2+(x+1)^2-x^2

Expanding each binomial product gives

(x^2+6x+9)-(x^2+4x+4)+(x^2+2x+1)-x^2=4x+6

Then if x=2001, we get

2004^2-2003^2+2002^2-2001^2=4\cdot2001+6=8010

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4 years ago
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Step-by-step explanation:

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PLEASE MARK BRAINLIEST!

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