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If a_(n) = sum_(r=0)^(n) (1)/(""^(n)C_(r)), then the value of (sumsum)_(0leiltjlen) (i/(""^(n)C_(i))+j/(""^(n)C_(j))) |
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Answer» `an^(2)` `= underset(0leiltjlen)(sumsum)((n-i)/(.^(n)C_(n-i))+(n-j)/(.^(n)C_(n-j)))` `= n underset(0leiltjlen)(sumsum)((1)/(.^(n)C_(i))+(1)/(.^(n)C_(j)))-S` `rArr S = n/2underset(0leiltjlen)(sumsum) ((1)/(.^(n)C_(i))+(1)/(.^(n)C_(j)))` `=n/2(underset(r=0)OVERSET(n-1)SUM(n-r)/(.^(n)C_(r))+underset(r=1)overset(n)sum(r)/(.^(n)C_(r)))` `= n/2(overset(n)underset(r=0)sum(n)/(.^(n)C_(r))) = (n^(2)a)/(2)` |
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