A) \[\frac{(2n+1)l}{3}\]
B) \[\frac{1}{n+1}\]
C) \[\frac{n({{n}^{2}}+1)l}{2}\]
D) \[\frac{2l}{n({{n}^{2}}+1)}\]
Correct Answer: A
Solution :
[a] \[{{x}_{cm}}=\frac{m\times \ell +2m\times 2\ell +3m\times 3\ell +...+nm\times n\ell }{m+2m+3m+..+nm}\] \[=\frac{(1+4+9+..+{{n}^{2}})\ell }{1+2+..+n}\] \[=\frac{\ell n(n+1)(2n+1)/6}{n(n+1)/2}=\frac{(2n+1)\ell }{3}\]
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