A) \[\frac{n(n-1)}{2}\]
B) \[\frac{n(n+2)}{2}\]
C) \[\frac{n(n+1)}{2}\]
D) \[\frac{(n-1)(n-2)}{2}\]
Correct Answer: C
Solution :
[c] :\[\frac{{{C}_{1}}}{{{C}_{0}}}+2.\frac{{{C}_{2}}}{{{C}_{1}}}+3.\frac{{{C}_{3}}}{{{C}_{2}}}+.....+n.\frac{{{C}_{n}}}{{{C}_{n-1}}}\] \[=\frac{n}{1}+2.\frac{n(n-1)/1.2}{n}+3\frac{n(n-1)(n-2)/3.2.1}{n(n-1)/1.2}+.....+n.\frac{1}{n}\]\[=n+(n-1)+(n-2)....+1=\sum\limits_{{}}^{{}}{n}=\frac{n(n+1)}{2}\]
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