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For and odd integer n ge 1, n^(3) - (n -...

For and odd integer `n ge 1, n^(3) - (n - 1)^(3) ` + ……
`+ (- 1)^(n-1) 1^(3)`

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To solve the given series for an odd integer \( n \geq 1 \): ### Step 1: Understand the series The series is given as: \[ S = n^3 - (n - 1)^3 + (n - 2)^3 - (n - 3)^3 + \ldots + (-1)^{n-1} \cdot 1^3 \] Since \( n \) is odd, the last term will be \( 1^3 \). ...
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Statement 1: Sum of the series 1^3-2^3+3^3-4^3++11^3=378. - Statement 2: For any odd integer ngeq1,n^3-(n-1)^3++(-1)^(n-1)1^3=1/4(2n-1)(n+1)^2dot

If n is an odd integer greater than or equal to 1, then the value of n^3 - (n-1)^3 + (n-1)^3 - (n-1)^3 + .... + (-1)^(n-1) 1^3

Knowledge Check

  • The sum of the squares of the first n positive integers 1^2 + 2^2 + 3^2 + ………+ n^2 is (n(n +1)(2n+1))/(6) . What is the sum of the squares of the first 9 positive integers?

    A
    90
    B
    125
    C
    200
    D
    285

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