on reading the following proof , I understand the reduction in the number of terms by using the "telescoping " method. My concern is terms such as the second term in:
$1\choose k+1$-$0\choose k+1$ which appears twice , is this a valid expression? as it =(1/((k+1)!.($0$-(k+1))!).Which in the second last term "looks " to be set to $0$
Working:
j+1 replacing n and k+1 replacing k in. ${n\choose k}$=${n-1\choose k-1}$+${n-1\choose k}$ gives. ${j+1\choose k+1}$=${j\choose k}$+${j\choose k+1}$ so:
${j\choose k}$=${j+1\choose k+1}$-${j\choose k+1}$
$\sum_{j=0}^n{j\choose k}$=$\sum_{j=0}^{n} [{j+1\choose k+1}- {j\choose k+1}$]=[$1\choose k+1$-$0\choose k+1$]+[$2\choose k+1$-$1\choose k+1$]+....+[$n+1\choose k+1$-$n\choose k+1$]=$n+1\choose k+1$-$0\choose k+1$=$n+1\choose k+1$
thanks ralph.
Hello, What I'm asking , is, $0\choose k+1$ what is the value of this expression , because in the second last term above it looks like the value is taken to be $0$
eg
$n+1\choose k+1$-$0\choose k+1$=$n+1\choose k+1$.But using the Combination formula 0! = 1 . So how is the expression arrived at ? thanks ralph
The problem is , the conventional formula for combinations: $n\choose k$ does not apply here:
For k $ >= $ 1,n=0, there are no k-element subsets of [0] = ∅, so,$0\choose k$ = 0
ralph