92 3U HSC Q6c)ii): Binomial Question⌄ (1 Viewer)

gamja · Jan 11, 2023

1992 3U HSC Q6c)ii)
Show that $1-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}-...+\left(-1\right)^{n}\frac{1}{n+1}\binom{n}{n}=\frac{1}{n+1}$

from $1+\binom{n}{1}x+\binom{n}{2}x^{2}+...+\binom{n}{n}x^{n}=\left(1+x\right)^{n}$ , I integrated both sides to get
$x+\frac{1}{2}\binom{n}{1}x^{2}+\frac{1}{3}\binom{n}{2}x^{3}+...+\frac{1}{n+1}\binom{n}{n}x^{n+1}=\frac{1}{n+1}\left(1+x\right)^{n+1}$

and then divided both sides by x to get
$1+\frac{1}{2}\binom{n}{1}x+\frac{1}{3}\binom{n}{2}x^{2}+...+\frac{1}{n+1}\binom{n}{n}x^{n}=\frac{\frac{1}{n+1}\left(1+x\right)^{n+1}}{x}$

and subbed in x=-1 to get LHS correct, but RHS became zero...

Could someone tell me where I went wrong? Thanks in advance!!

pikachu975 · Jan 11, 2023

So when you integrated you need the +C, so you have:

x + (1/2)(nC1)x^2 + ... = (1+x)^(n+1)/(n+1) + C
Sub in x = 0
C = -1/(n+1)

I'm not sure if you can divide both sides by x as x could be 0, but you could multiply both sides by x:

x^2 + (1/2)(nC1)x^3 + ... = x/(n+1) (1+x)^(n+1) - x/(n+1) = x/(n+1) ((1+x)^(n+1) - 1)

Now subbing x = -1:

LHS is correct still, RHS = -1/(n+1) (0 - 1) = 1/(n+1)

5uckerberg · Jan 11, 2023

gamja said:
1992 3U HSC Q6c)ii)
Show that $1-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}-...+\left(-1\right)^{n}\frac{1}{n+1}\binom{n}{n}=\frac{1}{n+1}$

from $1+\binom{n}{1}x+\binom{n}{2}x^{2}+...+\binom{n}{n}x^{n}=\left(1+x\right)^{n}$ , I integrated both sides to get
$x+\frac{1}{2}\binom{n}{1}x^{2}+\frac{1}{3}\binom{n}{2}x^{3}+...+\frac{1}{n+1}\binom{n}{n}x^{n+1}=\frac{1}{n+1}\left(1+x\right)^{n+1}$

and then divided both sides by x to get
$1+\frac{1}{2}\binom{n}{1}x+\frac{1}{3}\binom{n}{2}x^{2}+...+\frac{1}{n+1}\binom{n}{n}x^{n}=\frac{\frac{1}{n+1}\left(1+x\right)^{n+1}}{x}$

and subbed in x=-1 to get LHS correct, but RHS became zero...

Could someone tell me where I went wrong? Thanks in advance!!

I have a different take on this one, given the pattern of the binomial question you will have
$\left(1-x\right)^{n}=1-\begin{pmatrix}n\\1\end{pmatrix}x+\begin{pmatrix}n\\2\end{pmatrix}x^{2}+...+\begin{pmatrix}n\\n\end{pmatrix}\left(-x\right)^{n}$
Integrate that and you will have
$\frac{-\left(1-x\right)^{n+1}}{n+1}+C=x-\begin{pmatrix}n\\1\end{pmatrix}\frac{x^{2}}{2}+\begin{pmatrix}n\\2\end{pmatrix}\frac{x^{3}}{3}-...-\begin{pmatrix}n\\n\end{pmatrix}\frac{\left(-x\right)^{n+1}}{n+1}$

Here, $x=0,\;C=\frac{1}{n+1}$

$\frac{-\left(1-x\right)^{n+1}}{n+1}+\frac{1}{n+1}=x-\begin{pmatrix}n\\1\end{pmatrix}\frac{x^{2}}{2}+\begin{pmatrix}n\\2\end{pmatrix}\frac{x^{3}}{3}-...-\begin{pmatrix}n\\n\end{pmatrix}\frac{\left(-x\right)^{n+1}}{n+1}$

Divide by x and we will have
$-\frac{\left(1-x\right)^{n+1}}{x\left(n+1\right)}+\frac{1}{x\left(n+1\right)}=1-\begin{pmatrix}n\\1\end{pmatrix}\frac{x}{2}+\begin{pmatrix}n\\2\end{pmatrix}\frac{x^{2}}{3}-...-\begin{pmatrix}n\\n\end{pmatrix}\frac{-x\left(-x\right)^{n}}{x\left(n+1\right)}$

Which is equivalent to
$-\frac{\left(1-x\right)^{n+1}}{x\left(n+1\right)}+\frac{1}{x\left(n+1\right)}=1-\begin{pmatrix}n\\1\end{pmatrix}\frac{x}{2}+\begin{pmatrix}n\\2\end{pmatrix}\frac{x^{2}}{3}-...+\begin{pmatrix}n\\n\end{pmatrix}\frac{\left(-x\right)^{n}}{\left(n+1\right)}$

Next step sub in $x=1$

Then you have $0+\frac{1}{n+1}=1-\begin{pmatrix}n\\1\end{pmatrix}\frac{1}{2}+\begin{pmatrix}n\\2\end{pmatrix}\frac{1}{3}-...+\begin{pmatrix}n\\n\end{pmatrix}\frac{\left(-1\right)^{n}}{\left(n+1\right)}$

gamja · Jan 13, 2023

mx1 forum recently fr (thanks everyone <3)

92 3U HSC Q6c)ii): Binomial Question⌄ (1 Viewer)

gamja

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pikachu975

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5uckerberg

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gamja

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