i) Show that for all positivew integers n,
x[(1+x)n−1+(1+x)n−2+…(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≤k≤n
(n−1k−1)+(n−2k−1)+(n−3k−1)+⋯+(k−1k−1)=(nk)
I don't know how to do this second bit. Can someone help me please?
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
i) Show that for all positiveintegers n,
x[(1+x)n−1+(1+x)n−2+…(1+x)+1]=(1+x)n−1(1)
I can do this first bit no worries but then part 2 is
Hence show that for 1≤k≤n
(n−1k−1)+(n−2k−1)+(n−3k−1)+⋯+(k−1k−1)=(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)n−1=(n0)+(n1)x+(n2)x2⋯+(nk)xk⋯+(nn)xn−1∴The coefficient of xkis(nk)
LHS=x[(1+x)n−1+(1+x)n−2+…(1+x)+1]I want to find the coefficient of xkSo I am only interested in xk−1terms in the binomial expansionsx[(n−1k−1)xk−1+(n−2k−1)xk−1+⋯+(k−1k−1)xk−1]=[(n−1k−1)+(n−2k−1)+⋯+(k−1k−1)]xk
Equating coefficients we have
(n−1k−1)+(n−2k−1)+⋯+(k−1k−1)=(nk)
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