JEE MATHEMATICS MCQ: COMBINATIONS

1 Easy

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The value of

\sum_{\mathrm{r}=1}^{15} \mathrm{r}^{2}\left(\frac{{ }^{15} \mathrm{C}_{\mathrm{r}}}{{ }^{15} \mathrm{C}_{\mathrm{r}-1}}\right)

is equal to :

View answer and explanation

Correct option: D

680

Option D (680) is correct because after simplifying the summation expression and applying the formulas for the sum of first n natural numbers and the sum of squares of first n natural numbers, the result is 680.

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{\;^{n + 2}}{C_8}:{\;^{n - 2}}{P_4} = \frac{{57}}{{16}}

, then

n

is equal to