Question
Sum of the first $n$ cubes (alternative form): $(1+2+3+4+\cdots+n)^{2}$ Earlier we noted the formula for the sum of the first $n$ cubes was $\frac{n^{2}(n+1)^{2}}{4}$. An alternative is given by the formula shown.a. Verify the formula for $n=1,5,$ and 9b. Verify the formula using $1+2+3+\cdots+n=\frac{n(n+1)}{2}$
Step 1
For $n=1$, we have $S_{1}=1^{2}=1^{3}$, which verifies the formula. For $n=5$, we have $S_{5}=(1+2+3+4+5)^{2}=15^{2}=225$. The sum of the first five cubes is $1+8+27+64+125=225$, which verifies the formula. For $n=9$, we have Show more…
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