Prove that 3C1 - 7C2 - 11C3 - - - 4n - 1Cn - 1 - 2n - 1 - 2n

L.H.S = ∑_r = 1^n (4r 1) C_r = 4 #160; ∑_r = 1^n rC_r #160; ∑_r = 1^n C_r = 4 [ n 2 ^n 1 ] (2^ ...

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Question

Prove that
$3 C_{1} + 7 C_{2} + 11 C_{3} + . . . . + (4 n - 1) C_{n} = 1 + (2 n - 1) \cdot 2^{n}$

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

3 C_{1} + 7 C_{2} + 11 C^{3} + . . . . + (4 n - 1) C_{n}

The value of $C_{0} + 3 C_{1} + 5 C_{2} + 7 C_{3} + . . . . + (2 n + 1) C_{n}$ is equal to :

\frac{^{4 n} C_{2 n}}{^{2 n} C_{n}} = \frac{1.3.5...... (4 n - 1)}{{1.3.5...... (2 n - 1)}^{2}}

C_{0} + 3 C_{1} + 5 C_{2} + . . . . . + (2 n + 1) C_{n} = 32 t h e n \int_{- 1}^{1} (n^{2} - n + 1) d x =

^{4 n} C_{2 n} :^{2 n} C_{n} = 1.3.5... (4 n - 1) : [1.3.5... (2 n - 1)]^{2} .

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