The value of ∑r=115r^2(^15C_r)/(^15C_(r−1)) is equal to:

Correct answer: (D) 680. Given ∑ r=1 15 r^2 ( (^15 C_r)/(^15 C_(r−1)) ) Here (^15C_r)/(^15C_(r-1))=(16-r)/r ∑ r=1 15 r^2 ( (^15 C_r)/(^15 C_(r−1)) ) = ∑ r=1 15 r^2 (16-r)/r = ∑ r=1 15 r 16-r =16 ∑ r=1 15 r - ∑ r=1 15 r^2 =(16 ×15 ×16)/2-(15 ×31 ×16)/6 =8 ×15 ×16-5 ×8 ×31 =1920-1240=680

Q43 / 90 · JEE Main 2016 (09 Apr Online) · Mathematics

The value of $\sum_{r = 1}^{15} r^{2} (\frac{^{15} C_{r}}{^{15} C_{r - 1}})$ is equal to:

A
$1240$
B
$560$
C
$1085$
Correct answer
D
$680$

Solution

Given

\sum_{r = 1}^{15} r^{2} (\frac{^{15} C_{r}}{^{15} C_{r - 1}})

Here

\frac{^{15} C_{r}}{^{15} C_{r - 1}}

= \frac{16 - r}{r}

\sum_{r = 1}^{15} r^{2} (\frac{^{15} C_{r}}{^{15} C_{r - 1}})

= \sum_{r = 1}^{15} r^{2}

(\frac{16 - r}{r})

= \sum_{r = 1}^{15} r

(16 - r)

=

16 \sum_{r = 1}^{15} r - \sum_{r = 1}^{15} r^{2}

=

\frac{16 \times 15 \times 16}{2} - \frac{15 \times 31 \times 16}{6}

=

8 \times 15 \times 16 - 5 \times 8 \times 31

=

1920 - 1240 = 680

Hence, the correct answer is option: 'D'.

The value of ∑r=115r215Cr15Cr−1 is equal to:

Solution

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