frac{{{C_1}}}{{{C_0}}} + 2frac{{{C_2}}}{{{C_1 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) We know that

$\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + .... + n.\frac{{{C_n}}}{{{C_{n - 1}}}} = \frac{{n

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(b) We know that

$\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + .... + n.\frac{{{C_n}}}{{{C_{n - 1}}}} = \frac{{n(n + 1)}}{2}$

Putting $n=15$, then $\frac{{15 	imes (15 + 1)}}{2} = 120$.

$\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + .... + 15\frac{{{C_{15}}}}{{{C_{14}}}} = $

Similar Questions

$(1-x)^{101}\left(x^{2}+x+1\right)^{100}$ નાં વિસ્તરણમાં $x^{256}$ નો સહગુણક મેળવો.

જો ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$, તો ${C_0}{C_2} + {C_1}{C_3} + {C_2}{C_4} + {C_{n - 2}}{C_n}$= . . .

$\frac{1}{{1!(n - 1)\,!}} + \frac{1}{{3!(n - 3)!}} + \frac{1}{{5!(n - 5)!}} + .... = $

ધારોકે $\sum \limits_{r=0}^{2023} r^{2023} C_r=2023 \times \alpha \times 2^{2022}$, તો $\alpha$ નું મૂલ્ય $............$ છે.