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 :
\[\frac{{{C}_{1}}}{{{C}_{0}}}+2.\frac{{{C}_{2}}}{{{C}_{1}}}+3.\frac{{{C}_{3}}}{{{C}_{2}}}+.....+n.\frac{{{C}_{n}}}{{{C}_{n-1}}}\] \[=\frac{n}{1}+2\frac{n(n-1)/1.2}{n}+3\frac{n(n-1)(n-2)/3.2.1}{n(n-1)/1.2}+....+n.\frac{1}{n}\] \[=n+(n-1)+(n-2)....+1=\sum{n=\frac{n(n+1)}{2}}\] Trick: Put \[n=1,2,3\]....., then \[{{S}_{1}}=\frac{^{1}{{C}_{1}}}{^{1}{{C}_{0}}}=1\], \[{{S}_{2}}=\frac{^{2}{{C}_{1}}}{^{2}{{C}_{0}}}+2\frac{^{2}{{C}_{2}}}{^{2}{{C}_{1}}}=\frac{2}{1}+2.\frac{1}{2}=2+1=3\] By option, (put n=1,2......) (a) and (b) does not hold condition, but C \[\frac{n(n+1)}{2}\], put n =1, 2...... \[{{S}_{1}}=1,{{S}_{2}}=3\] which is correct.
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