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Use Mathematical Induction to prove that

$$1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \ldots + (n - 1) \cdot (n +1) = \dfrac{n(n-1)(2n + 5)}{6}$$ for all integers $n \geq 2$.

Base step is simple and true, but, I'm struggling to grasp the inductive step.

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Hint: Compute $$\dfrac{n(n-1)(2n+5)}{6} + n(n+2).$$

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HINT: In the second step we assume that formula

$$ (1 \cdot 3) + (2 \cdot 4) (3 \cdot 5) + \dots + (n - 1) \cdot (n +1) = \frac{n(n-1)(2n + 5)}{6} $$

holds for $n = k$ and from there demonstrate that it also holds for $n = k+1$:

\begin{align} (1 \cdot 3) + (2 \cdot 4) (3 \cdot 5) + \dots + (k - 1) \cdot (k +1) & = \dfrac{k(k-1)(2k + 5)}{6} \\ (1 \cdot 3) + (2 \cdot 4) (3 \cdot 5) + \dots + (k - 1) \cdot (k +1) + k \cdot (k + 2) & = \dfrac{k(k+1)(2k + 7)}{6} \end{align}

Since we know that: $$(1 \cdot 3) + (2 \cdot 4) (3 \cdot 5) + \dots + (k - 1) \cdot (k +1) = \dfrac{k(k-1)(2k + 5)}{6}$$ we can add $k\cdot(k+2)$ to both sides of the equation:

\begin{align} (1 \cdot 3) + (2 \cdot 4) (3 \cdot 5) + \dots + (k - 1) \cdot (k +1) + k \cdot (k + 2) & = \dfrac{k(k-1)(2k + 5)}{6} + k \cdot (k + 2) \\ & = \dfrac{k(k-1)(2k + 5) + 6 k \cdot (k + 2)}6 \\ & = \dfrac{k}{6}\big((k-1)(2k + 5) + 6(k+2)\big) \\ & = \dfrac{k}{6}\left(2k^2+9k+7\right ) \\ & = \dfrac{k(k+1)(2k + 7)}{6} \end{align}

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  • $\begingroup$ yeah but how is the final answer written highlighting that using mathmatical induction P(n) is indeed true for n>=2? $\endgroup$ – plushy May 25 '17 at 6:13
  • $\begingroup$ @plushy I don't think I understand your question. What is $P(n)$? Also, I pretty much wrote out the entire proof, what is the answer are you talking about? $\endgroup$ – user449440 May 25 '17 at 6:16
  • $\begingroup$ @plushy have you tried to see what happens with $n=0$ and $n=1$? $\endgroup$ – Joaquin Romera May 25 '17 at 19:34
  • $\begingroup$ Also try to write it as the sum $\sum_{i=?}^{n}(i-1)(i+1)$ $\endgroup$ – Joaquin Romera May 25 '17 at 19:36

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