By the principle of mathematical induction, prove that, for $n\ge 1$ $$1^2+3^2+5^2+\cdots+(2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}.$$

By the principle of mathematical induction, prove that, for n≥ 1 1²+3²+5²+…+(2n-1)² = (n(2n-1)(2n+1))/(3).

Exercise 4.4 · Q2

n\ge 1

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Let $P(n):1^2+3^2+\cdots+(2n-1)^2=\dfrac{n(2n-1)(2n+1)}3$ .

Step 1. Base case. $P(1)$ : LHS $=1$ ; RHS $=\dfrac{1\cdot1\cdot3}3=1$ . True.

Step 2. Inductive hypothesis. Assume $P(k):1^2+\cdots+(2k-1)^2=\dfrac{k(2k-1)(2k+1)}3$ .

Step 3. Inductive step. The $(k+1)$ th odd term is $(2(k+1)-1)^2=(2k+1)^2$ . $P(k+1)$ : $\dfrac{k(2k-1)(2k+1)}3+(2k+1)^2=\dfrac{(2k+1)\big[k(2k-1)+3(2k+1)\big]}3=\dfrac{(2k+1)(2k^2+5k+3)}3=\dfrac{(2k+1)(2k+3)(k+1)}3=\dfrac{(k+1)(2k+1)(2k+3)}3=\dfrac{(k+1)\big(2(k+1)-1\big)\big(2(k+1)+1\big)}3$ , exactly $P(k+1)$ .

Step 4. Conclusion. By PMI, $P(n)$ holds for all $n\ge1$ .

✓Final answer

n\ge1

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Q.By the principle of mathematical induction, prove that, for $n\ge 1$
[!FORMULA] $1^2+3^2+5^2+\cdots+(2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}.$

Concept understanding — Principle of Mathematical Induction

Q.By the principle of mathematical induction, prove that, for n≥1n\ge 1n≥1 [!FORMULA] 12+32+52+⋯+(2n−1)2=n(2n−1)(2n+1)3.1^2+3^2+5^2+\cdots+(2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}.12+32+52+⋯+(2n−1)2=3n(2n−1)(2n+1)​.

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Q.By the principle of mathematical induction, prove that, for $n\ge 1$
[!FORMULA] $1^2+3^2+5^2+\cdots+(2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}.$

Unlock everything free for 14 days