Question
If $${\left( {1 + x} \right)^{15}} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_{15}}{x^{15}}$$ then $${C_2} + 2{C_3} + 3{C_4} + ..... + 14{C_{15}}$$ is equal to
A.
$$14 \cdot {2^{14}}$$
B.
$${13 \cdot 2^{14}} + 1$$
C.
$${13 \cdot 2^{14}} - 1$$
D.
None of these
Answer :
$${13 \cdot 2^{14}} + 1$$
Solution :
The general term $${T_r} = \left( {r - 1} \right){\,^{15}}{C_r},r = 2,3,4,.....,15$$
$$\eqalign{
& \therefore {T_r} = r{\,^{15}}{C_r} - {\,^{15}}{C_r} \cr
& = 15 \cdot {\,^{14}}{C_{r - 1}} - {\,^{15}}{C_r}\left[ {\because r \cdot {\,^n}{C_r} = n \cdot {\,^{n - 1}}{C_{r - 1}}} \right] \cr
& \therefore \sum\limits_{r = 2}^{15} {{T_r} = 15\left[ {^{14}{C_1} + {\,^{14}}{C_2} + ..... + {\,^{14}}{C_{14}}} \right]} - \left[ {^{15}{C_2} + {\,^{15}}{C_3} + ..... + {\,^{15}}{C_{15}}} \right] \cr
& = 15\left[ {{2^{14}} - 1} \right] - \left[ {{2^{15}} - 1 - 15} \right] \cr
& = \left( {15 - 2} \right){2^{14}} + 1 = 13 \cdot {2^{14}} + 1 \cr} $$