Binomial Theorem Question (1 Viewer)

nightweaver066 · Sep 30, 2011

$\text{By assuming the basic result of}\\(1+x)^n = c_0 + c_1x + c_2x^2 + ...\text{ and } (1 - x)^n = c_0 - c_1x + c_2x^2 - c_3x^3 + ...\text{, prove that }\\ \frac{1}{2}c_1 + \frac{1}{4}c_3 + \frac{1}{6}c_5 + .... + \frac{1}{2n}c_{2n-1} = \frac{2^n -1}{n+1}$

I know i have to subtract one from the other, integrate and let x be a specific value but i can't seem to get the LHS $\frac{1}{2n}c_{2n-1}$ part.

Would be great if someone could post a solution with working out.

apollo1 · Sep 30, 2011

nightweaver066 said:
$\text{By assuming the basic result of}\\(1+x)^n = c_0 + c_1x + c_2x^2 + ...\text{ and } (1 - x)^n = c_0 - c_1x + c_2x^2 - c_3x^3 + ...\text{, prove that }\\ \frac{1}{2}c_1 + \frac{1}{4}c_3 + \frac{1}{6}c_5 + .... + \frac{1}{2n}c_{2n-1} = \frac{2^n -1}{n+1}$

I know i have to subtract one from the other, integrate and let x be a specific value but i can't seem to get the LHS $\frac{1}{2n}c_{2n-1}$ part.

Would be great if someone could post a solution with working out.

i think the 2n-1 part is in there to indicate that when you subtract the two the terms remaining have odd powers.

nightweaver066 · Sep 30, 2011

apollo1 said:
i think the 2n-1 part is in there to indicate that when you subtract the two the terms remaining have odd powers.

Mm.. Are you sure? I'm not meant to work through using the two expansions and change it in to that?

SpiralFlex · Sep 30, 2011

nightweaver066 said:
Mm.. Are you sure? I'm not meant to work through using the two expansions and change it in to that?

I got a similar answer to that of the proof by integrating over ranges from 0 to 1 and via subtraction. However I could not get the term you have said although I got everything else.

AAEldar · Sep 30, 2011

SpiralFlex said:
I got a similar answer to that of the proof by integrating over ranges from 0 to 1 and via subtraction. However I could not get the term you have said although I got everything else.

I tried as well and couldn't get it, I got very close but not exact >.<

SpiralFlex · Sep 30, 2011

The question does not make mathematical sense,

I don't see how $nC_{2n-1}$ is even possible.

SpiralFlex · Sep 30, 2011

I'm going to spend the next 3 hours on this problem and see if I can interpret it differently. I will be reporting back for duty very soon.

AAEldar · Sep 30, 2011

SpiralFlex said:
The question does not make mathematical sense,

I don't see how $nC_{2n-1}$ is even possible.

$\frac{n!}{(2n-1)!(-n+1)!}$

It's not.... D: Factorial doesn't apply to negatives.

And by the way Spiral, to get the proper $^nC_n$ notation you write it as ^nC_n. For future reference

SpiralFlex · Sep 30, 2011

AAEldar said:
$\frac{n!}{(2n-1)!(-n+1)!}$

It's not.... D: Factorial doesn't apply to negatives.

And by the way Spiral, to get the proper $^nC_n$ notation you write it as ^nC_n. For future reference

Thank you! So this question has a flaw in it somewhere. I will get onto it.

AAEldar · Sep 30, 2011

I have:

$2^n=2{^nC_1}+\frac{4}{n}{^nC_2}+\frac{6}{n}{^nC_3}+...+2{^nC_n}$

and

$\frac{2^n}{n+1}=\frac{1}{2}{^nC_0}+\frac{1}{2}{^nC_1}+\frac{1}{2}{^nC_2}+...+\frac{1}{2}{^nC_n}$

No idea.

taeyang · Sep 30, 2011

$\therefore \frac{2^{n+1}}{n+1} = \binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+...+\frac{1}{n+1}\binom{n}{n}....(1)\\\\ \therefore \frac{1}{n+1} = \binom{n}{0}-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}-...-\frac{1}{n+1}\binom{n}{n}....(2)\\\\ $Subtract (2) from (1)\\\\ \therefore \frac{2^{n+1}}{n+1}-\frac{1}{n+1} = \binom{n}{1}+\frac{1}{2}\binom{n}{3}+\frac{1}{3}\binom{n}{5}+....\frac{2}{n+1}\binom{n}{n}\\ \therefore \frac{2^n.2 -1}{n+1}=\binom{n}{1}+\frac{1}{2}\binom{n}{3}+\frac{1}{3}\binom{n}{5}+....\frac{2}{n+1}\binom{n}{n}\\$

I got that.. yeh.... probably no where near.

SpiralFlex · Sep 30, 2011

taeyang said:
$\therefore \frac{2^{n+1}}{n+1} = \binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+...+\frac{1}{n+1}\binom{n}{n}....(1)\\\\ \therefore \frac{1}{n+1} = \binom{n}{0}-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}-...-\frac{1}{n+1}\binom{n}{n}....(2)\\\\ $Subtract (2) from (1)\\\\ \therefore \frac{2^{n+1}}{n+1}-\frac{1}{n+1} = \binom{n}{1}+\frac{1}{2}\binom{n}{3}+\frac{1}{3}\binom{n}{5}+....\frac{2}{n+1}\binom{n}{n}\\ \therefore \frac{2^n.2 -1}{n+1}=\binom{n}{1}+\frac{1}{2}\binom{n}{3}+\frac{1}{3}\binom{n}{5}+....\frac{2}{n+1}\binom{n}{n}\\$

I got that.. yeh.... probably no where near.

I think mines closer. I will type it up.

nightweaver066 · Sep 30, 2011

I ended up with this..

$\frac{1}{2}c_1 + \frac{1}{4}c_3 + \frac{1}{6}c_5 + ... + c_n\frac{1 + (-1)^{n+1}}{2(n+1)} = \frac{2^n -1}{n+1}$

SpiralFlex · Sep 30, 2011

nightweaver066 said:
I ended up with this..

$\frac{1}{2}c_1 + \frac{1}{4}c_3 + \frac{1}{6}c_5 + ... + c_n\frac{1 + (-1)^{n+1}}{2(n+1)} = \frac{2^n -1}{n+1}$

Very similar to what I got. But I am sure there is a mistake to this question.

Drongoski · Sep 30, 2011

I think the question is a botched attempt to combine the following:

Show that:

For n even:

$\frac {2^n -1}{n+1} = \frac {1} {2}c_1+ \frac {1}{4}c_3 + \frac {1}{6}c_5 + \dots + \frac {1}{n}c_{n-1}$

For n odd:

$\frac {2^n - 1}{n+1} = \frac {1}{2}c_1 + \frac {1}{4}c_3 + \frac {1}{6}c_5 + \dots + \frac {1}{n+1}c_n$

So apollo1 made the correct observation.

or is this my botched attempt to explain the inconsistency.

Binomial Theorem Question (1 Viewer)

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