If ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + . . . . . . . + C n x n , then C 1 C 0 + 2 C 2 C 1 + 3 C 3 C 2 + . . . . . . . . + n C n C n − 1 = Options: A . &nbsp n ( n − 1 ) 2 B . &nbsp n ( n + 2 ) 2 C . &nbsp n ( n + 1 ) 2 D . &nbsp ( n − 1 ) ( n − 2 ) 2

Answer: Option C : C C 1 C 0 + 2 ⋅ C 2 C 1 + 3 ⋅ C 3 C 2 + . . . . . . . . + n ⋅ C n C n − 1 = n 1 + 2 n ( n − 1 ) 1.2 n + 3 n ( n − 1 ) ( n − 2 ) 3.2.1 n ( n − 1 ) 1.2 + . . . . . . + n . 1 n = n + ( n − 1 ) + ( n − 2 ) . . . . . . . . + 1 = ∑ n = n ( n + 1 ) 2

If (1+x)n=C0+C1x+C2x2+.......+Cnxn, then C1C0+2C2C1+3C3C2+........+nCnCnȡ

Discussion Forum : Binomial Theorem

Question -

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . + C_{n} x^{n}$ , then $\frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{3}}{C_{2}} + . . . . . . . . + \frac{n C_{n}}{C_{n - 1}} =$

Options:

A .

\frac{n (n - 1)}{2}

B .

\frac{n (n + 2)}{2}

C .

\frac{n (n + 1)}{2}

D .

\frac{(n - 1) (n - 2)}{2}

Answer: Option C
:
C

$\frac{C_{1}}{C_{0}} + 2 \cdot \frac{C_{2}}{C_{1}} + 3 \cdot \frac{C_{3}}{C_{2}} + . . . . . . . . + n \cdot \frac{C_{n}}{C_{n - 1}}$

$= \frac{n}{1} + 2 \frac{\frac{n (n - 1)}{1.2}}{n} + 3 \frac{\frac{n (n - 1) (n - 2)}{3.2.1}}{\frac{n (n - 1)}{1.2}} + . . . . . . + n . \frac{1}{n}$

$= n + (n - 1) + (n - 2) . . . . . . . . + 1 = \sum n = \frac{n (n + 1)}{2}$

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Discussion Forum : Binomial Theorem

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