Prove by induction that $\displaystyle\sum\limits_{k=m}^{\ n}{n\choose k}{k\choose m}={n\choose m}2^{n-m}$.
I can't figure out what is the base case. Could someone show the steps?
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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