Magic With Maths

Algebra Level 3

1 2 − 2 2 + 3 2 − 4 2 + . . . . . . + 9 9 2 − 10 0 2 = ? 1^{2}-2^{2}+3^{2}-4^{2}+......+99^{2}-100^{2} = ?


The answer is -5050.

Siva Prasad by Siva prasad , 20, India

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3 solutions

Arulx Z
Dec 12, 2015

We can use telescoping. By observing the first few elements of the series, we observe a pattern -

{ 1 }^{ 2 }-{ 2 }^{ 2 }=\left( 1-2 \right) \left( 1+2 \right) =-1-2\\ { 3 }^{ 2 }-{ 4 }^{ 2 }=\left( 3-4 \right) \left( 3+4 \right) =-3-4

Let's see if the pattern is consistent by proving it.

= n 2 − ( n + 1 ) 2 = ( n + n + 1 ) ( n − n − 1 ) = − 1 ( 2 n + 1 ) = − 2 n − 1 = − n + ( − n − 1 ) ={ n }^{ 2 }-{ \left( n+1 \right) }^{ 2 }\\ =\left( n+n+1 \right) \left( n-n-1 \right) \\ =-1\left( 2n+1 \right) \\ =-2n-1\\ =-n+\left( -n-1 \right)

Aha! Now we can use the pattern to simplify the series and write it as a AP.

= 1 2 − 2 2 + 3 2 − 4 2 + ⋯ + 9 9 2 − 10 0 2 = − 1 − 2 − 3 − 4 − ⋯ − 99 − 100 = − ( 1 + 2 + 3 + 4 + ⋯ + 99 + 100 ) =1^{ 2 }-2^{ 2 }+3^{ 2 }-4^{ 2 }+\dots +99^{ 2 }-100^{ 2 }\\ =-1-2-3-4-\dots -99-100\\ =-\left( 1+2+3+4+\dots +99+100 \right)

Now we can simplify it using the summation formula .

= − ( 1 + 2 + 3 + 4 + ⋯ + 99 + 100 ) = − ( 100 ⋅ 101 2 ) = − 5050 =-\left( 1+2+3+4+\dots +99+100 \right) \\ =-\left( \frac { 100\cdot 101 }{ 2 } \right) \\ =-5050

3 Helpful 1 Interesting 1 Brilliant 0 Confused

Moderator note:

Good approach of recognizing the difference of two squares.

Note a^2-b^2=(a-b)(a+b) the latter is difference of sqaurs

Mardokay Mosazghi - 5 years, 5 months ago

Solution https://www.youtube.com/watch?v=MJSPTF14OY4&ab_channel=RazingThunderRazingThunder

Razing Thunder - 2 weeks, 3 days ago

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Rwit Panda
Dec 3, 2015

It is easy to observe that if two terms are taken together, and a²-b²=(a+b)(a-b) is applied, the series can be collapsed to an easy arithmetic progression -3,-7,-11,….,199 with 50 terms.

Sum of this AP=50/2 [2(-3) + 49(-4)]=-5050. :)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Your solution is more difficult... Try this method...! :)

1 2 − 2 2 + 3 2 − 4 2 + . . . . . . + 9 9 2 − 10 0 2 1^{2}-2^{2}+3^{2}-4^{2}+......+99^{2}-100^{2}

The above is equal to ( 1 − 2 ) ( 1 + 2 ) + ( 3 − 4 ) ( 3 + 4 ) + . . . . . + ( 99 − 100 ) ( 99 + 100 ) (1-2)(1+2)+(3-4)(3+4)+.....+(99-100)(99+100)

That is equal to

( − 1 ) ( 1 + 2 ) + ( − 1 ) ( 3 + 4 ) + ( − 1 ) ( 5 + 6 ) + . . . . . . . + ( − 1 ) ( 99 + 100 ) (-1)(1+2)+(-1)(3+4)+(-1)(5+6)+.......+(-1)(99+100)

Take − 1 -1 outside..., then you will get it as − 1 × ( 1 + 2 + 3 + 4 + . . . . . . . . + 99 + 100 ) -1 \times (1+2+3+4+........+99+100)

Simply using n ( n + 1 ) 2 \frac { n(n+1) }{ 2 } , 1 + 2 + 3... + 99 + 100 = 5050 1+2+3...+99+100=5050

Then the answer will be − 1 × 5050 = − 5050 -1 \times 5050 = \boxed{-5050}

Simple...! :)

Siva prasad - 5 years, 6 months ago

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Ya, for a person not familiar with sum of terms in an arithmetic progression, your method is better.

But most of the people attempting this problem would be familiar with the same, it doesn't make any difference.

But still, thanks for the advice. I take it in good stead. :)

Rwit Panda - 5 years, 6 months ago

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