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# i) Binomial mathematics

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i)  Show that for all positivew integers n,

$$x[(1+x)^{n-1}+(1+x)^{n-2}+\dots (1+x)+1]=(1+x)^n-1\\$$

I can do this first bit no worries but then part 2 is

Hence show that for   $$1\le k\le n$$

$$\begin{pmatrix}n-1 \\k-1\end{pmatrix}+ \begin{pmatrix}n-2 \\k-1\end{pmatrix}+ \begin{pmatrix}n-3 \\k-1\end{pmatrix}+\dots + \begin{pmatrix}k-1 \\k-1\end{pmatrix}= \begin{pmatrix}n \\k\end{pmatrix}$$

I don't know how to do this second bit.  Can someone help me please?

Sep 7, 2016

#1
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Give n and k numerical value, which might be easier to visualize: i. e. n=11, k=3

(11 -1)C(3 - 1)=45 + (11 - 2)C(3 - 1)=36+ (11 - 3)C(3 - 1)=28+(11 - 4)C(3-1)=21........=11C3=165

45                   +         36                   +         28                 +       21........+15+ 10+6+3+1  =165

Sep 7, 2016
#2
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i)  Show that for all positiveintegers n,

$$x[(1+x)^{n-1}+(1+x)^{n-2}+\dots (1+x)+1]=(1+x)^n-1\qquad (1)\\$$

I can do this first bit no worries but then part 2 is

Hence show that for   $$1\le k\le n$$

$$\begin{pmatrix}n-1 \\k-1\end{pmatrix}+ \begin{pmatrix}n-2 \\k-1\end{pmatrix}+ \begin{pmatrix}n-3 \\k-1\end{pmatrix}+\dots + \begin{pmatrix}k-1 \\k-1\end{pmatrix}= \begin{pmatrix}n \\k\end{pmatrix}$$

I don't know how to do this second bit.  Can someone help me please?

Ok what you need to see is that is the coefficient of x^k  in the right hand side of equation 1 is   nCk

So I need to find the coefficient of x^k in the left hand side.

Those two coefficients mucs be equal

Looking at equation (1)

$$RHS\\ = (1+x)^n-1\\ =\begin{pmatrix}n\\0\end{pmatrix} +\begin{pmatrix}n\\1\end{pmatrix}x +\begin{pmatrix}n\\2\end{pmatrix}x^2\dots +\begin{pmatrix}n\\k\end{pmatrix}x^k\dots +\begin{pmatrix}n\\n\end{pmatrix}x^n-1\\ \therefore \mbox{The coefficient of }x^k \;\;is\;\;\begin{pmatrix} n\\k \end{pmatrix}$$

$$LHS=x[(1+x)^{n-1}+(1+x)^{n-2}+\dots (1+x)+1]\\ \mbox{I want to find the coefficient of }x^k\\ \mbox{So I am only interested in }x^{k-1} \;\;\mbox{terms in the binomial expansions}\\ x\left [\begin{pmatrix} n-1\\k-1 \end{pmatrix}x^{k-1}+\begin{pmatrix} n-2\\k-1 \end{pmatrix}x^{k-1}+\dots +\begin{pmatrix} k-1\\k-1 \end{pmatrix}x^{k-1} \right]\\ =\left [\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n-2\\k-1 \end{pmatrix}+\dots +\begin{pmatrix} k-1\\k-1 \end{pmatrix}\right]x^k\\$$

Equating coefficients we have

$$\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n-2\\k-1 \end{pmatrix}+\dots +\begin{pmatrix} k-1\\k-1 \end{pmatrix}= \begin{pmatrix} n\\k \end{pmatrix}$$

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Sep 8, 2016