Latest A website Aptitude Question SOLUTION: 0^2 - 1^2 + 2^2 - 3^2 + 4^2 - ..... - 99^2 + 100^2 = ?

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(#M40034718) A WEBSITE Numerical Ability question Sum of series Keep an EYE

0^2 - 1^2 + 2^2 - 3^2 + 4^2 - ..... - 99^2 + 100^2 = ?

Asked In A website Ann Theressa (10 years ago)

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RAKESH  (10 years ago)

0^2 - 1^2 + 2^2 - 3^2 + 4^2 - ..... - 99^2 + 100^2

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

= 2+1 + 4+3 +6+5 +... + 100+99

= 1+2+3+4+....+99+100

= 100*(100+1)/2

= 5050

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Bankim  (10 years ago)

0^2 - 1^2 + 2^2 - 3^2 + 4^2 - ..... - 99^2 + 100^2
= (2^2-1^2)+(4^2-3^2)+(6^2-5^2)+...+(100^2-99^2)
= 3+7+11+15+...195+199 total 50 terms
= (50/2)*(3+199)= 25*202=5050

Is this solution Helpfull? Yes (2) | No (3)  

vishnuvardhan  (10 years ago)

(0^2+2^2+4^2+_ _ +100^2)-(1^2+3^2+5^2+_ _ +99^2)
nth term of series (2n)^2-(2n-1)^2 [where n=50]
4n^2-4n^2+4n-1=4n-1
to take sum applying sigma or summation
4sigma n-sigma 1 [sigma 1 to n times is n]
4 sigma n-n
(4*n(n+1)/2)-n [taking n=50]
(4*50*51/2)-50=5050
sum of series is 5050

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