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For any odd integer n ≥ 1, n 3 – (n-1) 3 + ……….+ (-1) n-1 1 3 =…………

For any odd integer n ≥ 1, n3 – (n-1)3 + ……….+ (-1)n-1 13 =…………

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
7 years ago
Hello Student,
Please find the answer to your question
Ans. Since n is an odd integer, (-1)n-1 = 1 and n – 1, n – 3, n – 5, … are even integers.
We have
n3 + (n – 1)3 + (n – 2) 3 – (n – 3) 3 + …. + (-1)n – 1 13
= n3 + (n – 1) 3 + ( n – 2) 3 + ….. + 13 – 2[(n - 1) 3 + (n – 3) 3 + …… 23 ]
=[n3 + (n – 1)3 +( n – 2) 3 +……. + 13]
-2 * 23 [(n – 1/2) 3 + (n – 3/2) 3 + ….. + 13 ]
[ ∵ n – 1, n – 3 …………………….. are even integers]
Here the first square bracket contain the sum of cubes of Ist n natural numbers. Whereas the second square bracket contains the sum of the cubes of natural numbers from l to
(n – 1 /2), where n – 1, n – 3, ……………… are even integers. Using the formula for sum of cubes of lst n natural numbers we get the summation
= [ n (n + 1)/2]2 – 16 [ { 1/2 ( n – 1/2)( n – 1/2 + 1)]2
= 1/4n2 (n +1) 2 – 16 (n – 1) 2 ( n + 1) 2 /16 * 4
= 1/4 ( n + 1) 2 [n2 - ( n - 1) 2 = 1/4 ( n + 1) 2 (2n – 1)

Thanks
Deepak Patra
askIITians Faculty

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