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i)  Show that for all positivew integers n,

 

x[(1+x)n1+(1+x)n2+(1+x)+1]=(1+x)n1

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

 

Hence show that for   1kn

 

(n1k1)+(n2k1)+(n3k1)++(k1k1)=(nk)

 

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)n1+(1+x)n2+(1+x)+1]=(1+x)n1(1)

 

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

 

Hence show that for   1kn

 

(n1k1)+(n2k1)+(n3k1)++(k1k1)=(nk)

 

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)n1=(n0)+(n1)x+(n2)x2+(nk)xk+(nn)xn1The coefficient of xkis(nk)

 

 

LHS=x[(1+x)n1+(1+x)n2+(1+x)+1]I want to find the coefficient of xkSo I am only interested in xk1terms in the binomial expansionsx[(n1k1)xk1+(n2k1)xk1++(k1k1)xk1]=[(n1k1)+(n2k1)++(k1k1)]xk

 

Equating coefficients we have

 

(n1k1)+(n2k1)++(k1k1)=(nk)

 

*

 Sep 8, 2016

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