7.1 Q-19

Exercise 7.1 — Question 19

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): $\frac{1\cdot 2\cdot 3}{3}=\frac{6}{3}=2$

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

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Step 2: Inductive Hypothesis

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

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

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Step 3: Inductive Step

We must prove the statement is true for $n=k+1$, i.e., we need to show:

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

Starting from the LHS for $n=k+1$:

${\sum }_{j=1}^{k+1}j(j+1)={\sum }_{j=1}^{k}j(j+1)+(k+1)(k+2)$

Using the inductive hypothesis:

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

Factor out $(k+1)(k+2)$:

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

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

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

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

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Conclusion

By the Principle of Mathematical Induction, the statement

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

is true for all positive integers $n\in \mathbb{N}$.