Field hockey stick identity

Field hockey Stick Identity:.

$\sum_{i = 0}^{k} (i n + i) = (k n + k + 1)$
proof - negative binomial theorem eq4-1 - $1/ (1 - x)^{d + 1} = \sum_{n = 0}^{\infty} (d n + d) x^{n}$ - Partial sum of coefficients of formal power series eq4-2 - $i = 0 \sum k [x^{i}] F = [x^{k}] \frac{1}{1 - x} F$
- eq4-3
  - $(m n + m) = (n n + m)$
- $\sum_{i = 0}^{k} (i n + i) = \sum_{i = 0}^{k} [x^{i}] \sum_{m = 0}^{\infty} (m n + m) x^{m}$ … Partial sum of coefficients
- $\sum_{i = 0}^{k} [x^{i}] \sum_{m = 0}^{\infty} (m n + m) x^{m} = [x^{k}] \frac{1}{1 - x} \sum_{m = 0}^{\infty} (m n + m) x^{m}$ … by eq4-2
- $[x^{k}] \frac{1}{1 - x} \sum_{m = 0}^{\infty} (m n + m) x^{m} = [x^{k}] \frac{1}{1 - x} \sum_{m = 0}^{\infty} (n n + m) x^{m}$ … by eq4-3
- $[x^{k}] \frac{1}{1 - x} \sum_{m = 0}^{\infty} (n n + m) x^{m} = [x^{k}] \frac{1}{1 - x} \frac{1}{( 1 - x ) ^{n + 1}}$ … 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
- $[x^{k}] \frac{1}{( 1 - x ) ^{n + 2}} = [x^{k}] \sum_{m = 0}^{\infty} (n + 1 m + n + 1) x^{n}$ … by eq4-1
- $[x^{k}] \sum_{m = 0}^{\infty} (n + 1 m + n + 1) x^{n} = (n + 1 k + n + 1)$ … Coefficients of formal power series
- $(n + 1 k + n + 1) = (k k + n + 1)$ … by eq4-3
  - [Pascal’s triangle
  - Hockey Stick Identity | Math for Thinking Skills
binomial coefficient formula

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Field hockey stick identity

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