Mathematical Induction

[chemmer]

I really don't understand where to start on this problem. What I do understand is the general theory of induction; that you prove that the equation is true for n=1 and for n= k+1. However, I do not understand the logistics and thought process beyond this. Can someone please help?

Prove by mathematical induction:

\(\displaystyle \frac{1}{4} + 2\frac{1}{4} + 4\frac{1}{4} + ... + [2(n-1) + \frac{1}{4}] = \frac{4n^{2} + 3n}{4}\)

 

[soroban]

Hello, chemmer!

There is a typo . . . The right side should have "minus".

\(\displaystyle \text{Prove by mathematical induction: }\;\tfrac{1}{4} + 2\tfrac{1}{4} + 4\tfrac{1}{4} + \hdots + \left[2(n-1) + \tfrac{1}{4}\right] \;=\; \frac{4n^{2} - 3n}{4} \;=\;\frac{n(4n-3)}{4}\)

\(\displaystyle \text{Verify }S(1)\!:\quad \frac{1}{4} \:=\:\frac{1(4\cdot1 - 3)}{4}\quad\hdots\text{ True}\)

\(\displaystyle \text{Assume }S(k)\!:\quad \tfrac{1}{4} + 2\tfrac{1}{4} + 4\tfrac{1}{4} + \hdots + \left[2(n-1)+\tfrac{1}{4}\right] \;\;=\;\;\frac{k(4k-3)}{4}\) .[1]


\(\displaystyle \text{Now we must prove }S(k+1)\text{, which looks like this:}\)

. . \(\displaystyle \tfrac{1}{4} + 2\tfrac{1}{4} + 4\tfrac{1}{4} + \hdots + \left[2k + \tfrac{1}{4}\right] \;\;=\;\;\frac{(k+1)[4(k+1) - 3]}{4} \;=\;\frac{(k+1)(4k + 1)}{4}\) .[2]


\(\displaystyle \text{Add }\left(2k + \tfrac{1}{4})\text{ to both sides of [1]:}\)

. . \(\displaystyle \tfrac{1}{4} + 2\tfrac{1}{4} + 4\tfrac{1}{4} + \hdots + \left(2k + \tfrac{1}{4}\right) \;\;=\;\;\frac{k(4k-3)}{4} + 2k + \tfrac{1}{4} \;\;=\; \;\frac{4k^2 - 3k}{4} + \frac{8k}{4} + \frac{1}{4} \;\;=\;\;\frac{4k^2 + 5k + 1}{4}\)

. . \(\displaystyle \tfrac{1}{4} + 2\tfrac{1}{4} + 4\tfrac{1}{4} + \hdots + \left(2k + \tfrac{1}{4}\right) \;\;=\;\;\frac{(k+1)(4k+1)}{4}\)


\(\displaystyle \text{We have proved }[2]\;S(k+1)\quad\hdots\text{ The inductive proof is complete.}\)

 

[chrisr]

Hi chemmer,

you said that you weren't sure how the process of induction works.
I just want to add a little, though the procedure Soroban gave may be clear enough to you.

The given sum "appears to be" n(4n-3)/4.
If it is, then (n+1)(4[n+1]-3)/4 will sum the first n+1 terms.

This is (4n[sup:3ga5bd1y]2[/sup:3ga5bd1y]+4n-3n+4n+4-3)/4 = (4n[sup:3ga5bd1y]2[/sup:3ga5bd1y]+5n+1)/4.
If S[sub:3ga5bd1y]n[/sub:3ga5bd1y]+T[sub:3ga5bd1y]n+1[/sub:3ga5bd1y] is equal to the S[sub:3ga5bd1y]n+1[/sub:3ga5bd1y] above,
then S[sub:3ga5bd1y]2[/sub:3ga5bd1y]=S[sub:3ga5bd1y]1[/sub:3ga5bd1y]+T[sub:3ga5bd1y]2[/sub:3ga5bd1y], S[sub:3ga5bd1y]3[/sub:3ga5bd1y]=S[sub:3ga5bd1y]2[/sub:3ga5bd1y]+T[sub:3ga5bd1y]3[/sub:3ga5bd1y],
S[sub:3ga5bd1y]4[/sub:3ga5bd1y]=S[sub:3ga5bd1y]3[/sub:3ga5bd1y]+T[sub:3ga5bd1y]4[/sub:3ga5bd1y], S[sub:3ga5bd1y]5[/sub:3ga5bd1y]=S[sub:3ga5bd1y]4[/sub:3ga5bd1y]+T[sub:3ga5bd1y]5[/sub:3ga5bd1y] and on and on.

We check this by adding T[sub:3ga5bd1y]n+1[/sub:3ga5bd1y] to Sn,
n(4n-3)/4 + 2n+1/4 = (4n[sup:3ga5bd1y]2[/sup:3ga5bd1y]-3n+8n+1)/4 = (4n[sup:3ga5bd1y]2[/sup:3ga5bd1y]+5n+1)/4.
Therefore S[sub:3ga5bd1y]n[/sub:3ga5bd1y]+T[sub:3ga5bd1y]n+1[/sub:3ga5bd1y]=S[sub:3ga5bd1y]n+1[/sub:3ga5bd1y].

This now means that if S[sub:3ga5bd1y]1[/sub:3ga5bd1y]=1(4[1]-3)/4,
then S[sub:3ga5bd1y]2[/sub:3ga5bd1y] will be S[sub:3ga5bd1y]1[/sub:3ga5bd1y]+T[sub:3ga5bd1y]2[/sub:3ga5bd1y] which is n(4n-3)/4 with n=2,
S[sub:3ga5bd1y]3[/sub:3ga5bd1y] will be S[sub:3ga5bd1y]2[/sub:3ga5bd1y]+T[sub:3ga5bd1y]3[/sub:3ga5bd1y] which is n(4n-3)/4 with n=3,
and in general S[sub:3ga5bd1y]n[/sub:3ga5bd1y]=n(4n-3)/4 for the given series.