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hicaderasu1c8q 2022-06-04 Answered

1^{2} - 2^{2} + 3^{2} - 4^{2} + \dots + 1999^{2}

19992 using closed form formula for sum of squares of first n natural numbers

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opiftinkistj6mi6

Answered 2022-06-05 Author has 2 answers

If you first add all the squares:

1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + \dots

and then subtract the even squares:

- 2^{2} - 4^{2} - 6^{2} \dots

then you are left with only the odd squares:

1^{2} + 3^{2} + 5^{2} + \dots

You need to subtract the even squares one more time to get

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + \dots

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fulgoripb0b

Answered 2022-06-06 Author has 1 answers

We can use the identity

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

to tackle the sum in 2 simpler ways:
Way 1

\begin{aligned} 1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - . . . - 1998^{2} + 1999^{2} \\ = & 1 + (2 + 3) + (4 + 5) + \dots + (1998 + 1999) \\ = & 19990000 \end{aligned}

Way 2

\begin{aligned} (1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} . . . - 1998^{2}) + 1999^{2} \\ = & - ((1 + 2) + (3 + 4) + (5 + 6) + \dots + (1997 + 1998)) + 1999^{2} \\ = & 19990000 \end{aligned}

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