InterviewSolution
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Find the sum `(sumsum)_(0leiltjlen) jxx""^(n)C_(i)`. |
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Answer» `underset(0leiltjlen)(sumsum)j..^(n)C_(i)` `= underset(r=0)overset(n-1)sum.^(n)C_(r)[(r+1)+(r+2)+"...."+(n)]` `= underset(r=0)overset(n-1)sum.^(n)C_(r)[(r+1)+(r+2)+"...."(r+(n-r))]` `= underset(r=0)overset(n-1)sum.^(n)C_(r)(n-r)/(2)(r+1+n)` `= 1/2 underset(r=0)overset(n)sum.^(n)C_(r) (n(n+1)-r-r^(2))` ` = 1/2 [n(n+1)underset(r=0)overset(n)sum.^(n)C_(r)-underset(r=0)overset(n)sumr^(n)C_(r)-underset(r=0)overset(n)sumr^(2)..^(n)C_(r)]` `=1/2[n(n+1).2^(n)-n underset(r=0)overset(n)sum.^(n-1)C_(r-1)-n underset(r=0)overset(n)sumr..^(n-1)C_(r-1)]` `=1/2[n(n+1).2^(n)-n.2^(n-1)-n underset(r=0)overset(n)sum((n-1)..^(n-2)C_(r-2)+.^(n+1)C_(r-1))]` `= 1/2[n(2n+1).2^(n-1)-n(n-1).2^(n-2)-n.2^(n-1)]` `= n(3n+1).2^(n-3)` |
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