If n- 2 then 3- C 1 -4- C 2 -5- C 3 - -1 n-1 n-2 - C n is equal to

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Higher Order Derivatives

If n≥ 2 the...

Question

If $n \geq 2$ then $3. C_{1} - 4. C_{2} + 5. C_{3} - . . . . . . + {(- 1)}^{n - 1} (n + 2) . C_{n}$ is equal to

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Similar questions

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

= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{n}

n \geq 2

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

\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)}

S = \frac{2}{1} {^{n} C}_{0} + \frac{2^{2}}{2} {^{n} C}_{1} + \frac{2^{3}}{3} {^{n} C}_{2} + . . . . . + \frac{2^{n + 1}}{n + 1} {^{n} C}_{n}

S

(1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{n},

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

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