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Exercise 4.4 · Q1

Q.By the principle of mathematical induction, prove that, for n≥1n\ge 1
[!FORMULA] 13+23+33+⋯+n3=(n(n+1)2)21^3+2^3+3^3+\cdots+n^3 = \left(\dfrac{n(n+1)}{2}\right)^2

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✓ Free question

Let P(n):13+23+⋯+n3=(n(n+1)2)2P(n):1^3+2^3+\cdots+n^3=\left(\dfrac{n(n+1)}2\right)^2.

Step 1. Base case. P(1)P(1): LHS =13=1=1^3=1; RHS =(1⋅22)2=12=1=\left(\dfrac{1\cdot2}2\right)^2=1^2=1. True.

Step 2. Inductive hypothesis. Assume P(k):13+⋯+k3=(k(k+1)2)2P(k):1^3+\cdots+k^3=\left(\dfrac{k(k+1)}2\right)^2.

Step 3. Inductive step. P(k+1)P(k+1): 13+⋯+k3+(k+1)3=(k(k+1)2)2+(k+1)3=k2(k+1)2+4(k+1)34=(k+1)2(k2+4(k+1))4=(k+1)2(k+2)24=((k+1)(k+2)2)21^3+\cdots+k^3+(k+1)^3=\left(\dfrac{k(k+1)}2\right)^2+(k+1)^3=\dfrac{k^2(k+1)^2+4(k+1)^3}4=\dfrac{(k+1)^2\big(k^2+4(k+1)\big)}4=\dfrac{(k+1)^2(k+2)^2}4=\left(\dfrac{(k+1)(k+2)}2\right)^2, which is exactly P(k+1)P(k+1).

Step 4. Conclusion. By PMI, P(n)P(n) holds for all n≥1n\ge1.

✓Final answer

Proved for all n≥1n\ge1 by the Principle of Mathematical Induction.

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