Statement Prove by induction that ∑i=1ni3=(n(n+1)2)2\displaystyle\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2i=1∑ni3=(2n(n+1))2. Solution Base (n=1n=1n=1): 13=11^3=113=1 and (1⋅22)2=1\left(\dfrac{1\cdot 2}{2}\right)^2=1(21⋅2)2=1. True. Step: suppose ∑i=1ki3=(k(k+1)2)2\displaystyle\sum_{i=1}^k i^3=\left(\dfrac{k(k+1)}{2}\right)^2i=1∑ki3=(2k(k+1))2 and add (k+1)3(k+1)^3(k+1)3: (k(k+1)2)2+(k+1)3=(k+1)2[k24+(k+1)]=(k+1)2 k2+4k+44=((k+1)(k+2)2)2,\left(\frac{k(k+1)}{2}\right)^2+(k+1)^3 = (k+1)^2\left[\frac{k^2}{4}+(k+1)\right] = (k+1)^2\,\frac{k^2+4k+4}{4} = \left(\frac{(k+1)(k+2)}{2}\right)^2,(2k(k+1))2+(k+1)3=(k+1)2[4k2+(k+1)]=(k+1)24k2+4k+4=(2(k+1)(k+2))2, which is the formula for n=k+1n=k+1n=k+1. By induction the claim holds for every n≥1n\ge 1n≥1. ∎ Links Topics: Set theory Concepts: Principle of induction Methods: Induction Skills: Proving Exercise type: Proof