Field hockey stick identity

Field hockey Stick Identity:.

• $\sum_{i=0}^k \binom{n+i}{i} = \binom{n+k+1}{k}$

• proof - negative binomial theorem eq4-1 - $1/(1-x)^{d+1} = \sum_{n=0}^\infty \binom{n+d}{d}x^n$ - Partial sum of coefficients of formal power series eq4-2 - $\displaystyle \sum_{i=0}^{k} [x^{i}]F = [x^{k}] \frac{1}{1-x}F$

  • eq4-3

    • $\binom{n + m}{m} = \binom{n + m}{n}$

  • $\sum_{i=0}^k \binom{n+i}{i} = \sum_{i=0}^{k} [x^{i}] \sum_{m=0}^\infty \binom{n+m}{m}x^m$ ... Partial sum of coefficients

  • $\sum_{i=0}^{k} [x^{i}] \sum_{m=0}^\infty \binom{n+m}{m}x^m = [x^{k}] \frac{1}{1-x}\sum_{m=0}^\infty \binom{n+m}{m}x^m$ ... by eq4-2

  • $[x^{k}] \frac{1}{1-x}\sum_{m=0}^\infty \binom{n+m}{m}x^m = [x^{k}] \frac{1}{1-x}\sum_{m=0}^\infty \binom{n+m}{n}x^m$ ... by eq4-3

  • $[x^{k}] \frac{1}{1-x}\sum_{m=0}^\infty \binom{n+m}{n}x^m = [x^{k}] \frac{1}{1-x} \frac{1}{(1-x)^{n+1}} $ ... by eq4-1

  • $[x^{k}] \frac{1}{1-x} \frac{1}{(1-x)^{n+1}} = [x^{k}] \frac{1}{(1-x)^{n+2}}$ ... Sorting out

  • $[x^{k}] \frac{1}{(1-x)^{n+2}} = [x^{k}] \sum_{m=0}^\infty \binom{m+n+1}{n+1}x^n$ ... by eq4-1

  • $[x^{k}] \sum_{m=0}^\infty \binom{m+n+1}{n+1}x^n = \binom{k+n+1}{n+1}$ ... Coefficients of formal power series

  • $\binom{k+n+1}{n+1} = \binom{k+n+1}{k}$ ... by eq4-3

    • [Pascal's triangle
    • 
    • Hockey Stick Identity | Math for Thinking Skills

• binomial coefficient formula

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