Prove that C0 - C12 - C23 - C34 - - - 1n - 1

Adding the two results of #160; we get , 2^n + 1 1 + 1n + 1 = 2 ( C_0 + C_23 + C_45 + ... ) or #160; 2^nn + 1 = ( C ...

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Particular Solution of a Differential Equation

Prove that ...

Question

Prove that $C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \frac{C_{3}}{4} + . . . . . . = \frac{1}{n + 1}$

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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_{3}}{5} . . . . . . = \frac{1}{(n + 1) (n + 2)}

C_{0} + 2. C_{1} + 4. C_{2} + 8. C_{3} + . . . . . . + 2^{n} . C_{n} = 3^{n}

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

r = 0, 1, 2, . . . ., n

C_{0} \cdot C_{r} + C_{1} \cdot C_{r + 1} + C_{2} \cdot C_{r + 2} + . . . . + C_{n - r} \cdot C_{n} =^{2 n} C_{(n + r)}

C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . + C_{n}^{2} =^{2 n} C_{n}

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

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