Proof by induction: 2 + 5 + 8 + ... + (3n - 1) = [n(3n+1)]/2

[kimberlyd1020]

Use induction to show that, for all positive integers n,

2+5+8+...+(3n-1) = n(3n+1)/2

 

[soroban]

Re: Proof by induction

Hello, kimberlyd1020!

\(\displaystyle \text{Prove by induction: }\;2+5+8+ \hdots +(3n-1) \:= \:\frac{n(3n+1)}{2}\)

\(\displaystyle \text{Verify }S(1)\!:\quad 2 \:=\:\frac{1(3\!\cdot\!1+1)}{2} \:=\:2\quad\hdots \text{ True}\)

\(\displaystyle \text{Assume }S(k)\!:\quad 2 + 5 + 8 + \hdots + (3k-1) \:=\:\frac{k(3k+1)}{2}\)


\(\displaystyle \text{Add }3(k+1)-1 \:=\:3k+2\text{ to both sides: }\;\)
. . \(\displaystyle \underbrace{2 + 5 + 8 + \hdots + (3k-1) + (3k+2)} \;=\;\frac{k(3k+1)}{2} + (3k+2)\)

\(\displaystyle \text{The left side is the left side of }S(k+1)\)
\(\displaystyle \text{We must show that the right side is the right side of }S(k+1)\!:\;\;\frac{(k+1)(3[k+1]+1)}{2}\)


\(\displaystyle \text{The right side is: }\;\frac{k(3k+1)}{2} + (3k+2) \;=\;\frac{k(3k+1) + 2(3k+2)}{2} \;=\;\frac{3k^2 + 7k+4}{2}\)

. . \(\displaystyle \text{which simplifies to: }\;\frac{(k+1)(3k+4)}{2} \;=\;\underbrace{\frac{(k+1)(3[k+1] + 1)}{2}}_{\text{Right side of }S(k+1)}\)
The inductive proof is complete!