SOLUTION: (a) Compute the sum:101^2 - 97^2 + 93^2 - 89^2 + ...+ 5^2 - 1^2. (b) Compute the sum (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + ... + (a+d)^2 - a^2. Can y

Algebra ->  Sequences-and-series -> SOLUTION: (a) Compute the sum:101^2 - 97^2 + 93^2 - 89^2 + ...+ 5^2 - 1^2. (b) Compute the sum (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + ... + (a+d)^2 - a^2. Can y      Log On


   

Question 1168361: (a) Compute the sum:101^2 - 97^2 + 93^2 - 89^2 + ...+ 5^2 - 1^2.
(b)
Compute the sum (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + ... + (a+d)^2 - a^2.
Can you please explain in detail?
Answer by ikleyn(52792) About Me  (Show Source):
You can put this solution on YOUR website!
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Compute the sum: 101^2 - 97^2 + 93^2 - 89^2 + ...+ 5^2 - 1^2.
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SUM = 101^2 - 97^2 + 93^2 - 89^2 + ...+ 5^2 - 1^2 = group the term in pairs =

    101^2 - 97^2 +

  +  93^2 - 89^2 + 

  +  85^2 - 81^2 +

  . . . . . . . . . . .

  +   5^2 - 1^2.


To each pair (= to each short line) apply the identity  a^2 - b^2 = (a+b)*(a-b).

Notice that the second factor in the right side of this identity is always equal to 4.

Therefore, you get


SUM = 4*((101+97) + (93+89) + (85+81) + . . . + (5+1)) = 

    = 4*( 101+97  +  93+89  +  85+81  + . . . +  5+1 )


The sum in parentheses is the sum of the arithmetic progression with the first term of 1, the common difference of 4
and the last term of 101.


The number of terms in this progression is  %28101-1%29%2F4+%2B+1 = 26.

Therefore the sum in parentheses is  %28%28101%2B1%29%2F2%29%2A26 = 1326.


Hence, the original sum is  SUM = 4*1326 = 5304.


ANSWER.  The requested sum is  5304.

Solved.

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Regarding the other problem, I think that the same idea works there, too.