Some explaination would be helpful
if$\quad$ $a^n = a(a - h)....[a - (n - 1)h ]$ $\quad$ and $\quad$ $a^0=1$ $\quad$ then $\qquad$ prove $\quad$ $(a + b)^n = \sum_{m=0}^n C_n^m a^{n - m} b^{m}$,
Some explaination would be helpful
if$\quad$ $a^n = a(a - h)....[a - (n - 1)h ]$ $\quad$ and $\quad$ $a^0=1$ $\quad$ then $\qquad$ prove $\quad$ $(a + b)^n = \sum_{m=0}^n C_n^m a^{n - m} b^{m}$,
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