Statistics And Probability Question 360
Question: The weighted mean of first n natural numbers whose weights are equal to the squares of corresponding numbers is
[Pb. CET 1989]
Options:
A) $ \frac{n+1}{2} $
B) $ \frac{3n(n+1)}{2(2n+1)} $
C) $ \frac{(n+1)(2n+1)}{6} $
D) $ \frac{n(n+1)}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Weighted mean = $ \frac{{{1.1}^{2}}+{{2.2}^{2}}+……+n.n^{2}}{1^{2}+2^{2}+……+n^{2}} $ $ =\frac{\Sigma n^{3}}{\Sigma n^{2}}=\frac{\frac{n(n+1)}{2}\frac{n(n+1)}{2}}{\frac{n(n+1)(2n+1)}{6}} $ $ =\frac{3n(n+1)}{2(2n+1)} $ .
BETA
