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Greatest Binomial Coefficients

If c0, c1, ...

Question

If $c_{0}, c_{1}, c_{2}, . . . . . c_{n}$ denote the coefficients in the expansion of $(1 x)^{n}$ , prove that $c_{1} + 2 c_{2} + 3 c_{3} + . . . . . . + n c_{n} = n {.2}^{n - 1}$ .

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Solution

We know,
$\Rightarrow$ $(1 + x)^{n} = C_{0} + x C_{1} + x^{2} c_{2} + . . . + x^{n} C_{n}$
On differentiating both sides w.r.t. $x,$ we get,
$\Rightarrow$ $n (1 + x)^{n - 1} = C_{1} + 2 x C_{2} + . . . + n x^{n - 1} C_{n}$
Put $x = 1$ we get
$\Rightarrow$ $n {.2}^{n - 1} = C_{1} + 2 C_{2} + 3 C_{3} + . . . . + n C_{n}$ ---- Hence proved.

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C_{0}, C_{1}, C_{2}, . . . . . . . C_{n}

denote the coefficients in the expansion of

{(1 + x)}^{n}

C_{1} + 2 C_{2} + 3 C_{3} + . . . . n C_{n} = n . {(2)}^{n - 1}

c_{0}, c_{1}, c_{2}, . . . . c_{n}

denote the coefficients in the expansion of

(1 + x)^{n}

\frac{c_{1}}{c_{0}} + \frac{2 c_{2}}{c_{1}} + \frac{3 c_{3}}{c_{2}} + . . . . . + \frac{n c_{n}}{c_{n - 1}} = \frac{n (n + 1)}{2}

c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}

denote the coefficients in the expansion of

{(1 + x)}^{n}

\frac{c_{1}}{c_{0}} + \frac{2 c_{2}}{c_{1}} + \frac{3 c_{3}}{c_{2}} + . . . \frac{n c_{n}}{c_{n - 1}} = \frac{n (n + 1)}{2}

C_{0}, C_{1}, C_{2} . . . ., C_{n}

denote the binomial coefficients in the expansion of

{(1 + x)}^{n}

\frac{C 1}{C 0} + 2 \frac{C 2}{C 1} + + 3 \frac{C 3}{C 2} + . . . . . + n \frac{C n}{C n - 1}

C_{0}, C_{1}, C_{2}, . . . C_{n}

are the binomial coefficients in the expansion of

(1 + x)^{n}

then prove 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}

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