Alternating sum of squares

Notice that : $10^2-9^2=(10+9)(10-9)=19$

$8^2-7^2=(8+7)(8-7)=15$

$6^2-5^2=(6+5)(6-5)=11$

$4^2-3^2=(4+3)(4-3)=7$

$2^2-1^2=(2+1)(2-1)=3$ .Therefore the sum $S=19+15+11+7+3$ which is an arithmetic progression with $d=4$ , $a_1=3$ and $n=5$ .We can use the formula :

$S=\frac{n}{2}(2a_1+(n-1)d)$ , and by substituting we get $S=55$

i think that the easiesr solution is

10+9+8+7+6+5+4+3+2+1

if u still don't understand u can comment :)

10^2 − 9^2 +8^ 2 −7^ 2 +6^ 2 −5^ 2 +4^ 2 −3^ 2 +2^ 2 −1^ 2 ?

= (10 + 9)(10 – 9) + (8 +7)(8 – 7) + (6 + 5)(6 – 5) + (4 + 3)(4 – 3) + (2 + 1)(2 – 1)

= (10 + 9) + (8 +7) + (6 + 5) + (4 + 3) + (2 + 1)

= sum of numbers from 1 to 10 = 10 × 11/2 = 55

A square of a number is the number multiplied by itself. You can figure it out either by figuring out all of the squares and adding and subtracting the numbers as the question says, or you can add up all of the numbers without including the 2's for the squares. 10+9+8+7+6+5+4+3+2+1=55.

$a^{2}$ - $b^{2}$ = ( a + b ) (a - b)

Here every ( a - b ) will be 1

So every two term goes to :

( 10 + 9 ) + ( 8 + 7 ) + ( 6 + 5 ) + ( 4 + 3 ) + ( 2 + 1 )

=n (n + 1) / 2

=10 (10 + 1) / 2

=55

4^2 - 3^2 + 2^2 - 1^2 =

(4+3)(4-3) + (2+1)(2-1) =

4 + 3 + 2 + 1

That way the answer is:

10+9+8+7+6+5+4+3+2+1 =

55

10^2-9^2+8^2-7^2+6^2-5^2+4^2-3^2+2^2-1^2 simplify each to get.. 100-81+64-49+36-25+16-9+4-1 then solve according to the properties given Ans: 55

100-81+64-49+36-25+16-9+3=19+15+11+7+3=55

=100-81+64-49+36-25+16-9+4-1 =55

For two consecutive numbers square difference is the sum of numbers like (10 * 10) - (9 * 9) = 10 + 9. This will solve the problem quick.

100-81+64-49+36-25+16-9+4-1=55

using the formula a^2 - b^2 =(a+b)(a-b) we get (10+9)(1)+(8+7)(1)+(6+5)(1)+(4+3)(1)+(2+1)(1) which simplifies to 10+9+8+7+6+5+4+3+2+1

To get the sum of the series we can count it or we can use the formula (n)(n+1)/2 which'll again simplify to 10(10+1)/2 which is equal to ,the solution,55

S=( (1) ^ 2 + (2) ^2 + ............. (10) ^ 2 ) - ( 2* ( (2) ^2 ( 1^2 + 2^2 + 3^2 + 4^2 + 5^2) ) )
what we have to find is (-S)

100-81+64-49+36-25+16-9+2-1=55

(10^2-9^2)=,100-81=19, (8^2-7^2)=,64-49=15, (6^2-5^2)=,36-25=9, (4^2-3^2)=,16-9=7, (2^2-1^2)=,4-1=3. so finally answer is 19+15+9+7+3=55.

19+15+11+7+3=55

simply addition and subtraction

there is pattern forming in this question-

10^2 - 9^2 =19

8^2 - 7^2 =15 ...... & So on

Therefore a series is forming with difference 4 -

19+15+11+7+3=55

sum of numbers 1 to 10 = 55

a^2-b^2=(a+b)(a-b).calculate all 10^2-9^2=19,8^2-7^2=15,6^2-5^2=11,4^2-3^2=7,2^2-1^2=3.then 19+15+11+7+3=55.

(10-9)(10+9) + (8-7)(8+7) + .........+(2-1)(2+1) = 19+15+11+7+3 = 55

TAKE the positive and negative values separately. Add positive values. from negative values take negative sign as common. and solve it.

100+64+36+16+4 = 220 81+49+25+9+1 =165 less 220-165 = 55 ans

(10 + 9)(10 - 9) + (8 +7)(8-7) +(6+5)(6-5) + (4+3)(4-3) +(2+1)(2-1) or 10+9+8+7+6+5+4+3+2+1

with calculator :D

100-81+64-49+36-25+16-9+4-1 = 55

55

100-81+64-49+36-25+16-9+4-1=55

Note that $10^{2}-9^{2} = 10 + 9$ , $8^{2}-7^{2} = 8 + 7$ ...

Hence

$10^{2}-9^{2}+8^{2}-...-3^{2}+2^{2}-1^{2}$

$=10 + 9 + 8 + ... + 3 + 2 + 1$

$= \frac{10 \times 11}{2} = \boxed{55}$ .

10^2 - 1^2 = 99 -9^2 + 2^2 = -77 8^2 - 3^2 = 55 -7^2 + 4^2 = -33 6^2 - 5^2 = 11 55 is in the middle...answer is 55...i just figured that there is a pattern..

100-81+64-49+36-25+16-9+4-1=55

seperate the positive terms & also the negative terms.... after the above step we have 220 (solving +terms) & -165 ( negative terms) ...........=55

100-91+64-49+36-25+16-9+4-1=55

100-81+64-49+36-25+16-9+4-1=55

55 is the correct answer

19+15+11+7+3=55

(100)-(81)+(64)-(49)+(36)-(25)+(16)-(9)+(4)-(1)=55

Just take the squares and solve!

(10+9)(10-9) + (8+7)(8-7) +...........+ (2+1)(2-1) =1+2+3+4+5+6+7+8+9+10 =55

(2+1)(2-1)+(4+3)(4-3)+(6+5)(6-5)+(8+7)(8-7)+(10+9)(10-9)=3+7+11+15+19=55

simply , this is not a bog series so we can calculate easily sum even number squares are 210 and odd are 155 210-155=55 u can use series formula

=100-81+64-49+36-25+16-9+4-1 =55

Getting the squares of the numbers given ,100-99+64-49+36-25+16-9+4-1 rearrange the terms.100+64+36+16+4-99-49-25-1-9.adding positive numbers we get 220.by adding negative numbers we get 165.By subtraction, 220-165=55