7.1 Q-29

Exercise 7.1 — Question 29

Prove by Mathematical Induction:

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

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Step 1: Base Case ($n=1$)

Left-hand side (LHS): $1\cdot (1+1)=1\cdot 2=2$

Right-hand side (RHS): $\frac{1(1+1)(1+2)}{3}=\frac{1\cdot 2\cdot 3}{3}=2$

Since LHS $=$ RHS $=2$, the statement is true for $n=1$.

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Step 2: Induction Hypothesis

Assume the statement is true for $n=m$, i.e., assume:

${\sum }_{k=1}^{m}k(k+1)=\frac{m(m+1)(m+2)}{3}$

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Step 3: Inductive Step ($n=m+1$)

We need to prove: ${\sum }_{k=1}^{m+1}k(k+1)=\frac{(m+1)(m+2)(m+3)}{3}$

Starting from the LHS: ${\sum }_{k=1}^{m+1}k(k+1)={\sum }_{k=1}^{m}k(k+1)+(m+1)(m+2)$

Using the induction hypothesis: $=\frac{m(m+1)(m+2)}{3}+(m+1)(m+2)$

Factor out $(m+1)(m+2)$: $=(m+1)(m+2)\left(\frac{m}{3}+1\right)$

$=(m+1)(m+2)\cdot \frac{m+3}{3}$

$=\frac{(m+1)(m+2)(m+3)}{3}$

This is exactly the RHS for $n=m+1$.

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Conclusion

Since the statement holds for $n=1$ (base case), and assuming it holds for $n=m$ implies it holds for $n=m+1$ (inductive step), by the Principle of Mathematical Induction the statement is true for all positive integers $n$.

$\boxed{\sum _{k=1}^{n}k(k+1)=\frac{n(n+1)(n+2)}{3}}$