Mathematical Induction problem.

1. Apr 27, 2012

charmedbeauty

1. The problem statement, all variables and given/known data

(1/2!)+(2/3!)+(3/4!)+.....+(n/(n+1)!)

a) calculate for a few small values of n.
b) Make a conjecture about a formula for this expression
c)Prove your conjecture by mathematical induction.

2. Relevant equations

3. The attempt at a solution

So for the first part I just used values n=1,2,3

so..

n=1

1/2!=1/2

n=2

2/3! = 1/6+1/2

n=3

3/4! = 1/8+1/6+1/2.

for part b)

make a conjecture about the formula

It should be ((n+1)!-1)/(n+1)!

for part c) I am getting stuck...

test for n=1, which is true

assume true for n=k

show true for n=k+1

so....

(1/2!)+(2/3!+(3/4!)+....+(k/(k+1)!)+((k+1)/(k+2)!) = ((k+2)!-1)/(k+2)!

where

(1/2!)+(2/3!+(3/4!)+....+(k/(k+1)!)= ((k+1)!-1)/(k+1)!

so...

((k+1)!-1)/(k+1)!+(k+1)/(k+2)!= ((k+2)!-1)/(k+2)!

but Im lost as to where to go from here, have I made a mistake???

Help!

Last edited: Apr 27, 2012
2. Apr 27, 2012

Office_Shredder

Staff Emeritus
2/3! is 1/3, not 1/6. Witing it out as 1/8+1/3+1/2 (correcting your answer for n=3) is not particularly enlightening. Put the fractions together!

n=1: 1/2
n=2: 5/6
n=3 47/48

is some sort of pattern beginning to emerge?

Your conjectured formula is just the definition so isn't what they're looking for. What you should try to do is come up with what's called a "closed form" expression - an expression for the sum that involves no big summation of terms.

3. Apr 27, 2012

charmedbeauty

How about now is that correct for b)??(look at original post I edited). but I'm a little lost on c)