A) \[\frac{3n(n+1)}{2(2n+1)}\]
B) \[\frac{2n+1}{3}\]
C) \[\frac{3n+1}{4}\]
D) \[\frac{3n+1}{2}\]
Correct Answer: C
Solution :
| [c] Here for each \[{{x}_{i}}=i,\] |
| weight \[{{w}_{i}}={{i}^{2}}+i\] |
| Hence, the required mean |
| \[=\frac{\sum{{{w}_{i}}{{x}_{i}}}}{\sum{{{w}_{i}}}}=\frac{\sum\limits_{i=1}^{n}{i({{i}^{2}}+i)}}{\sum\limits_{i=1}^{n}{({{i}^{2}}+i)}}\] |
| \[=\frac{\sum\limits_{i=1}^{n}{{{i}^{3}}}+\sum\limits_{i=1}^{n}{{{i}^{2}}}}{\sum\limits_{i=1}^{n}{{{i}^{2}}}+\sum\limits_{i=1}^{n}{i}}\] |
| \[=\frac{\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}+\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}\] |
| \[=\frac{\frac{n(n+1)}{2}\left\{ \frac{n(n+1)}{2}+\frac{2n+1}{3} \right\}}{\frac{n(n+1)}{2}\left\{ \frac{2n+1}{3}+1 \right\}}\] |
| \[=\frac{3{{n}^{2}}+7n+2}{2(2n+4)}=\frac{(3n+1)(n+2)}{4(n+2)}=\frac{3n+1}{4}\] |
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