A) \[\frac{n(n-1)}{2}\]
B) \[\frac{{{n}^{2}}(n-1)}{4}\]
C) \[\frac{n(n+1)}{4}\]
D) \[\frac{n(n+1)}{2}\]
Correct Answer: C
Solution :
[c] : Mean square deviation about the value x= 0 is \[\frac{1}{N}\sum\limits_{i=1}^{n}{{{f}_{i}}x_{i}^{2}=\frac{1}{{{2}^{n}}}\sum\limits_{r=0}^{n}{{{r}^{2n}}{{C}_{r}}}}\][where, \[N=\sum\limits_{{}}^{{}}{f]}\] \[=\frac{1}{{{n}^{n}}}n.\sum\limits_{r=1}^{n}{r{{.}^{n-1}}{{C}_{r-1}}}=\frac{1}{{{2}^{n}}}n\sum\limits_{r=1}^{n}{{{[(r-1)+1]}^{n-1}}{{C}_{r-1}}}\] \[=\frac{1}{{{n}^{n}}}.n\left\{ \sum\limits_{r=1}^{n}{{{(r-1)}^{n-1}}{{C}_{r-1}}}+\sum\limits_{r=1}^{n}{^{n-1}{{C}_{r-1}}} \right\}\] \[=\frac{1}{{{2}^{n}}}.n\left\{ (n-1)\sum\limits_{r=2}^{n}{^{n-2}{{C}_{r-2}}+{{2}^{n-1}}} \right\}\] \[=\frac{1}{{{2}^{n}}}n\left\{ (n-1){{2}^{n-2}}+{{2}^{n-1}} \right\}\] \[=\frac{n(n+1){{2}^{n-2}}}{{{2}^{n}}}=\frac{n(n+1)}{4}\]
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