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I have had more than a day of difficulty with this problem. I have watched many youtube videos on induction and have not been able to solve this problem particularly. I would appreciate helpful hints. Heres the question:

Prove: $$\forall n\gt 1, 1*2+2*3+3*4+...+n(n+1)=\frac{(n(n+1)(n+2))}{3}$$

I have shown my work below. I would appreciate some hints and guidance.

My Attempt:

$$1*2+2*3+3*4+...+n(n+1)=\frac{(n(n+1)(n+2))}{3}$$ Base case: $n=1$ $$1(1+1)=\frac{(1(1+1)(1+2))}{3}$$ $$2=\frac{6}{3}=2$$

The base case works.

Induction Hypothesis: $n=k$

$$1*2+2*3+3*4+...+k(k+1)=\frac{(k(k+1)(k+2))}{3}$$

Assume the Induction Hypothesis is true for $k$.

Then, we show: $$\frac{(k(k+1)(k+2))}{3}+(k+1)=\frac{k^2+6k+3}{3}$$

I get pretty confused trying to apply the inductive hypothesis here because it is difficult to treat the $n(n+1)$ as the $n$ since it involves multiplication.

Again. I would appreciate hints or even an explantation of this problem, since I have exercised my resources on YouTube and reading the text for well over a day now.

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The inductive step consists in showing that $$\frac{n(n+1)(n+2)}3+(n+1)(n+2)=\frac{(n+1)(n+2)(n+3)}3,$$ which shouldn't be too hard.

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You look great right up through here.

Assume: $1\cdot2+2\cdot3+3\cdot4+...+k(k+1)=\frac{(k(k+1)(k+2))}{3}$

We must show that:

$1\cdot2+2\cdot3+3\cdot4+...+k(k+1)+(k+1)(k+2)=\frac{(k+1)(k+2)(k+3)}{3}$

$1\cdot2+2\cdot3+3\cdot4+...+k(k+1)+(k+1)(k+2) = \frac{(k(k+1)(k+2))}{3}+(k+1)(k+2)\ $ by the inductive hypothesis.

Rather than multiply everything out, notice that $(k+1)(k+2)$ is a common factor.

$(k+1)(k+2)(\frac{k}{3} + 1)\\ \frac{(k+1)(k+2)(k + 3)}{3}\\ $

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Inductive Hypothesis:-

Assume $P(k)$ is true

$1\cdot2+2\cdot3+3\cdot4+.....+\sum _{ n=1 }^{ k }{ n(n+1) } =\frac{k(k+1)(k+2)}{3}$

Now show that Inductive Hypothesis is true. Show that $P(k+1)$ is true $$\sum _{ n=1 }^{ k+1 }{ n(n+1) } =\frac{k+1(k+1+1)(k+1+2)}{3}=\frac{k+1(k+2)(k+3)}{3}$$ Now, $$\sum _{ n=1 }^{ k+1 }{ n(n+1) }=\sum _{ n=1 }^{ k }{ n(n+1) }++(k+1)(k+1+1)$$ $$=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$ $$=\frac{k\color{red}{(k+1)(k+2)}+3\color{red}{(k+1)(k+2)}}{3}$$ $$=\frac{(k+1)(k+2)}{3}[k+3]$$ $$=\frac{(k+1)(k+2)(k+3)}{3}$$

By the principle of mathematical induction $1\cdot2+2\cdot3+3\cdot4+.....+n(n+1)=\frac{n(n+1)(n+2)}{3}$

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  • $\begingroup$ I factored out $(k+1)(k+2)$, then we get $(k+1)(k+2)\frac{k+3}{3}$ $\endgroup$ – tien lee Jun 7 '18 at 2:55

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