7.1 Q-16

Exercise 7.1 — Question 16

Prove by Mathematical Induction that for all positive integers $n$:

$1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}$

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

Left-hand side (LHS): $1^2=1$

Right-hand side (RHS): $\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{1\cdot 2\cdot 3}{6}=\frac{6}{6}=1$

Since LHS $=$ RHS $=1$, 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:

$1^2+2^2+3^2+\cdots +k^2=\frac{k(k+1)(2k+1)}{6}$

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

We need to prove that:

$1^2+2^2+\cdots +k^2+(k+1{)}^2=\frac{(k+1)(k+2)(2k+3)}{6}$

Starting from the LHS and using the inductive hypothesis:

$\underset{=\frac{k(k+1)(2k+1)}{6}}{\underbrace{1^2+2^2+\cdots +k^2}}+(k+1{)}^2$

$=\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2$

Factor out $(k+1)$:

$=(k+1)\left[\frac{k(2k+1)}{6}+(k+1)\right]$

$=(k+1)\left[\frac{k(2k+1)+6(k+1)}{6}\right]$

Expand the numerator:

$k(2k+1)+6(k+1)=2k^2+k+6k+6=2k^2+7k+6$

Factorise $2k^2+7k+6$:

$2k^2+7k+6=(k+2)(2k+3)$

Therefore:

$=\frac{(k+1)(k+2)(2k+3)}{6}$

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

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Conclusion

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

$\boxed{1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}}$