If $${C_0},{C_1},{C_2},.....,{C_{15}}$$ are binomial coefficients in $${\left( {1 + x} \right)^{15}},\,$$ then $$\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + ..... + 15\frac{{{C_{15}}}}{{{C_{14}}}} = $$
A.
60
B.
120
C.
64
D.
124
Answer :
120
Solution :
General term of the given series is
$$r\frac{{^n{C_r}}}{{^n{C_{r - 1}}}} = n + 1 - r$$
By taking summation over $$n,$$ we get
$$\eqalign{
& \sum\limits_1^{15} {r\frac{{^n{C_r}}}{{^n{C_{r - 1}}}}} = \sum\limits_{n = 1}^{15} {\left( {n + 1 - r} \right)} = \sum\limits_1^{15} {\left( {16 - r} \right)} \cr
& = 16 \times 15 - \frac{1}{2} \cdot 15 \times 16 \cr} $$
By using sum of $$n$$ natural numbers $$ = \frac{{n\left( {n + 1} \right)}}{2}$$
$$= 240 - 120 = 120$$
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is