For any odd integer n >= 1,n^3 - (n - 1)^3 + ... ... n + ( - 1)^n - 11^3 = ... ...

Since n is an odd integer (-1)^n-1=1, n-3, n-5, ... . are even integers Let S_n=n^3-(n-1)^3+(n-2)^3-(n-3)^3+ * s .-2^3+1^3nAdding and subtracting terms, we writenn n S_n=n^3+(n-1)^3+(n-2)^3+(n-3)^3+2^3+1^3 n -2[(n-1)^3+(n-3)^3+ * s+2^3] n =[1^3+2^3+ * s+n^3]-2[2^3+4^3+ * s+(n-1)^3] n =[1^3+2^3+ * s+n^3]-2 * 2^3[1^3+2^3+ * s+((n-1 / 2))^3] n =(n^2(n+1)^2 / 4)-frac16[((n-1 / 2))^2((n+1 / 2))^2]4 n =(1 / 4) n^2(n+1)^2-(1 / 4)(n-1)^2(n+1)^2 n =(1 / 4)(n+1)^2[n^2-(n-1)^2]=(1 / 4)(n+1)^2(2 n-1)n n

For any odd integer n ? 1,n^{3} - (n - 1)^{3} + + (... | Sequence & Series (Mathematics)

Since

n

is an odd integer

(-1)^{n-1}=1, n-3, n-5, \ldots

. are even integers Let

S_n=n^3-(n-1)^3+(n-2)^3-(n-3)^3+\cdots .-2^3+1^3

\nAdding and subtracting terms, we write\n

\n\begin{aligned}\n& S_n=n^3+(n-1)^3+(n-2)^3+(n-3)^3+2^3+1^3 \\\n& -2\left[(n-1)^3+(n-3)^3+\cdots+2^3\right] \\\n& =\left[1^3+2^3+\cdots+n^3\right]-2\left[2^3+4^3+\cdots+(n-1)^3\right] \\\n& =\left[1^3+2^3+\cdots+n^3\right]-2 \cdot 2^3\left[1^3+2^3+\cdots+\left(\frac{n-1}{2}\right)^3\right] \\\n& =\frac{n^2(n+1)^2}{4}-\frac{16\left[\left(\frac{n-1}{2}\right)^2\left(\frac{n+1}{2}\right)^2\right]}{4} \\\n& =\frac{1}{4} n^2(n+1)^2-\frac{1}{4}(n-1)^2(n+1)^2 \\\n& =\frac{1}{4}(n+1)^2\left[n^2-(n-1)^2\right]=\frac{1}{4}(n+1)^2(2 n-1)\n\end{aligned}\n

Explanation

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