\sum_{r=0}^{n-1}{}\sum_{j=1}^{n-r}{}(n-j)Cr=?
a.1
b.2^{n}
c.2^{n}-1
d.2^{n-1}
-
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3 Answers
1luk, i feel the lower limit for the first summation, canot be 0, becoz, we possibly cannot have (n-1)Cn, can we.., so i doubt the correctnes of the question..