Question
Show that $C_1+C_2+C_3+\ldots . .+C_6=63$

Answer

Since $C_0+C_1+C_2+C_3+\ldots . .+C_n=2^n$
Putting n = 6, we get
$C_0+C_1+C_2+\ldots . .+C_6=2^6$
$\therefore C_0+C_1+C_2+\ldots \ldots+C_6=64$
But, $C_0=1$
$\therefore 1+C_1+C_2+\ldots . .+C_6=64$
$\therefore C_1+C_2+\ldots . .+C_6=64-1=63$

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