Prove the following statements with either induction, strong induction or proof by smallest counterexample. Prove that $(1+2+3+\cdots+n)^{2}=1^{3}+2^{3}+3^{3}+\cdots+n^{3}$ for every $n \in \mathbb{N}$ .

Solution

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$\text{\underline{\textcolor{#c34632}{Remember}}}$ : $1+2+3+\cdots +k=\dfrac{k(k+1)}{2}$

$\textbf{Proof by Mathematical induction}$

For $n=1$ : $(1)^2=1=1^3=1$ thus it is true for $n=1$ .

Assume it is true for $n=k$ then $\color{#4257b2}(1+2+3+\cdots +k)^2=1^3+2^3+3^3+\cdots +k^3 \qquad \cdots (1)$

Now for $n=k+1$ :

\begin{align*} &(1+2+3+\cdots +k+k+1)^2\\ &=[(1+2+3+\cdots +k)+(k+1)]\\ &=(1+2+3+\cdots +k)^2+2(1+2+3+\cdots +k)(k+1)+(k+1)^2\\ &\overset{{\color{#4257b2}(1)}}{=}1^3+2^3+3^3+\cdots +k^3 +(k+1)[2(1+2+3+\cdots+k)+k+1]\\ &=1^3+2^3+3^3+\cdots +k^3 +(k+1)[\cancel{2}\dfrac{k(k+1)}{\cancel{2}}+(k+1)]\\ &=1^3+2^3+3^3+\cdots +k^3+(k+1)^2(k+1) \\ &=1^3+2^3+3^3+\cdots +k^3+(k+1)^3 \end{align*}

Thus it is true for $n=k+1$ .

Therefore by mathematical induction

(1+2+3+\cdots +k)^2=1^3+2^3+3^3+\cdots +k^3

for every integer $n \in \Bbb N$ . $\qquad \;\; \blacksquare$

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