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Definition of Functions

prove that ;-...

Question

prove that ;-
$^{n} c_{1} + 2 \cdot^{n} c_{2} + 3 \cdot^{n} c_{3} + . . . . + n \cdot^{n} c_{n} = n \cdot 2^{n - 1}$

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Solution

Let $(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} +^{n} C_{2} x^{2} + . . . . . . .^{n} C_{n} x^{n}$
Differentiate both sides w.r.t x
$n (1 + x)^{n - 1} =^{n} C_{1} +^{2 n} C_{2} x +^{3 n} C_{3} x^{2} + . . . . . . . (n)^{n} C_{n}$
Put $x = 1$
$n (2)^{n - 1} =^{n} C_{1} + {2.} n C_{2} + 3.^{n} C_{3} + . . . . n .^{n} C_{n}$
Hence proved

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C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \frac{C_{3}}{4} + . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{(n + 1)}

\frac{C_{0}}{2} + \frac{C_{1}}{3} + \frac{C_{2}}{4} . . . . . . + \frac{C_{n}}{n + 2} = \frac{1 + n \cdot 2^{n + 1}}{(n + 1) (n + 2)}

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

Q. If n is a positive integer then prove that

C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + . . . . . . . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

n \times 2^{n - 1} =^{n} C_{1} + 2 \times^{n} C_{2} + . . . . . n \times^{n} C_{n} .

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