Find the value of $1996^2-1995^2 + 1994^2-1993^2 + \dots + 2^2-1^2.$

Question

Find the value of $$1996^2-1995^2 + 1994^2-1993^2 + \dots + 2^2-1^2.$$

What is wrong with the following reasoning. By the difference of squares formula we have that the sum can be broken into $$(1996-1995)(1996+1995) + (1994-1993)(1994+1993) + \dots + (2-1)(2+1)$$

but all the terms with minus signs are $1$ so we are left with $$1996+1995+1994+1993+\dots+2+1 = \frac{1996(1997)}{2}=1993006$$ which is incorrect as the correct answer should be $998000$.

Though I wonder how the sum on factorization ends up with $(2-1)(2+1)$, i.e. $3$, yet you go down to $1$ in the next line, and still end up with the correct answer despite this seeming to imply you should be $3$ off. — PrincessEev, Jul 12 '23 at 20:49
That was probably the intended solution. Looks like a problem for school students to practice the difference of squares and sum of first $n$ natural numbers as Gauss did. — Petar, Jul 12 '23 at 20:50
This was a problem in a UK math competition book for school students and they provide the solution as $$\sum_{n=1}^{998} ((2n)^2 - (2n-1)^2) = \frac{4(998)(998+1)}{2} - 998 = 998000$$ @Peter — Louie, Jul 12 '23 at 20:54
Well ... $\frac{4(998)(998+1)}{2}-998=1993006$ actually. They miscalculated the last expression. — , Jul 12 '23 at 21:02
Yeah. The last expression is $\approx 2\times 1000\times 1000 = 2,000,000,\ $ not $\approx 1,000,000.$ — Adam Rubinson, Jul 12 '23 at 21:28
It's simply the difference of the sum of even and odd squares. There are formulas that give the sum of odd and even squares. Once you have the sum of each, you can calculate the difference. — user25406, Jul 13 '23 at 10:57

Mostafa · Answer 1 · 2023-07-13T01:03:35.430

-1

$$1996^2-1995^2 + 1994^2-1993^2 + \dots + 2^2-1^2＝\sum_{k＝1}^{998}\left({(2k)^2-(2k-1)^2}\right)\\~\\＝\sum_{k＝1}^{998}\left({4k-1}\right)＝4\sum_{k＝1}^{998}k\;-\sum_{k＝1}^{998}1＝\frac {4(998)(999)}{2}-998＝1993006$$

edited Jul 13 '23 at 01:03

answered Jul 13 '23 at 00:08

Mostafa

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better switch $k$ to $n$ or vice-versa. – peek-a-boo Jul 13 '23 at 01:00
Yes , of course – Mostafa Jul 13 '23 at 01:02

Find the value of $1996^2-1995^2 + 1994^2-1993^2 + \dots + 2^2-1^2.$

1 Answers1