If {(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) $\frac{{{C_1}}}{{{C_0}}} + 2.\frac{{{C_2}}}{{{C_1}}} + 3.\frac{{{C_3}}}{{{C_2}}} + ..... + n.\frac{{{C_n}}}{{{C_{n - 1}}}}$

$ = \frac{n}{1} + 2\frac{{n(n - 1)/1.2}}{n} + 3\frac{{n(n - 1)(n - 2)/3.2.1}}{{n(n - 1)/1.2}} + .... + n.\frac{1}{n}$

$ = n + (n - 1) + (n - 2).... + 1 = \sum {n = \frac{{n(n + 1)}}{2}} $

Trick : Put $n = 1,2,3$....., then ${S_1} = \frac{{^1{C_1}}}{{^1{C_0}}} = 1$

,${S_2} = \frac{{^2{C_1}}}{{^2{C_0}}} + 2\frac{{^2{C_2}}}{{^2{C_1}}} = \frac{2}{1} + 2.\frac{1}{2} = 2 + 1 = 3$

By option, (put $n=1,2.....$.) $(a)$ and $(b)$ does not hold condition, but $(c)$ $\frac{{n(n + 1)}}{2}$,

put $n =1, 2...$... ${S_1} = 1,{S_2} = 3$ which is correct.

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}}}} = $

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