help with induction proof for 1(1x2) + 1/(2x3) + ...

[gmc]

i am having a tough time understanding how to do inductions. could someone help me with this problem?

Find a formula for 1/(1×2)+1/(2×3)+⋯+1/(n(n+1)) by examining the values of this expression for small values of n. Also prove the formula you conjectured in part a.

 

[soroban]

Re: help with inductions

Hello, gmc!

Edit: Wow! .I don't know what happened to my formatting!

(a) Find a formula for: .\(\displaystyle \dfrac{1}{1\cdot2} + \dfrac{1}{2\cdot3} + \dfrac{1}{3\cdot 4} + \cdots + \dfrac{1}{n(n+1)}\)
. . .by examining the value of the expression for small value of \(\displaystyle n.\)

(b) Prove the formula you conjectured in part (a).

(a) Let's do what they suggested . . .

\(\displaystyle \begin{array}{cccc}n = 1: & \dfrac{1}{1\cdot2} &=&\dfrac{1}{2} \\ \\
n=2: & \dfrac{1}{1\cdot2} + \dfrac{1}{2\cdot3} &=& \dfrac{2}{3} \\ \\
n=3: & \dfrac{1}{1\cdot2} + \dfrac{1}{2\cdot3} + \dfrac{1}{3\cdot4} &=& \dfrac{3}{4} \\ \\
n=4: & \dfrac{1}{1\cdot2} + \dfrac{1}{2\cdot3} + \dfrac{1}{3\cdot4} + \dfrac{1}{4\cdot5} &=& \dfrac{4}{5} \end{array}\)


It looks like: .\(\displaystyle \:\dfrac{1}{1\cdot2} + \dfrac{1}{2\cdot3} + \dfrac{1}{3\cdot 4} + \hdots + \dfrac{1}{n(n+1)} \:=\:\dfrac{n}{n+1}\)



(b) Now use Mathematical Induction to prove this conjecture.

[COLOR=#e00e]. . [/COLOR]\(\displaystyle \displaystyle S(n) \:=\:\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)} \:=\:\frac{n}{n+1}\)

\(\displaystyle \displaystyle \text{Verify S(1): }\:S(1) \:=\:\frac{1}{1\cdot 2} \:=\:\frac{1}{2} \:\hdots\text{ True.}\)

\(\displaystyle \displaystyle \text{Assume S(k): }\:S(k) \:=\:\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \hdots + \frac{1}{k(k+1)} \:=\:\frac{k}{k+1}\)


\(\displaystyle \text{Add }\frac{1}{(k+1)(k+2)}\text{ to both sides:}\)

. . \(\displaystyle \displaystyle \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} \;=\;\) .\(\displaystyle \dfrac{k}{k+1} + \dfrac{1}{(k+1)(k+2)} \)

\(\displaystyle \text{The right side is: }\:\dfrac{k(k+2)}{(k+1)(k+2)} + \dfrac{1}{(x+1)(k+2)} \:=\: \dfrac{k^2+2k + 1}{(k+1)(k+2)}\)

. . . . . . . . . . . . . . \(\displaystyle =\; \dfrac{(k+1)^2}{(k+1)(k+2)} \;=\;\dfrac{k+1}{k+2} \)


We have proved \(\displaystyle S(k\!+\!1)\) . . . The inductive proof is complete.

 

[georgeburninator]

It looks like the previous post covered the pattern, so I'll start addressing the proof!

Since the first entry in the pattern is n = 1 (and you should double-check, maybe it works for n = 0?), you'll want to use that first value n = 1 as your base case in the proof.

For the inductive step, substitute k+1 for n. How can you pull out the pattern for up to k out of the k+1 pattern. If you can isolate it, then you can use your inductive hypothesis to simplify the pattern through k. Then it's a matter of showing that the simplified term plus all of the pattern leftover works out to the formula substituting k+1.

If you're struggling with the actual structure of induction, I would recommend looking at www.inductiveproofs.com. I believe they also have a similar example to your question in their examples.