Prove that 3n - C2- 3n - 2 - C4- 3n - 4 - - 2n - 11 - 2n

L.H.S 12 [ (3 + 1)^n + (3 1)^n ] = 12 [ 4^n #160; + 2^n ]=2^n 1 [ 2^n + 1 ] ...

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Multiplication of Any Polynomial

Prove that ...

Question

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

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lim n \to \infty \frac{3^{n} - 2^{n}}{3^{n} + 2^{n}}

(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}

lim n \to \infty \frac{3^{n + 1} + 2^{n + 2}}{3^{n - 1} + 2^{n - 2}}

If f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + x^{n} & (1 - x)^{n} & 2 + x^{n} (2 + x)^{n} & (2 + x)^{n} & 1 (3 - x)^{n} & 1 & 3 + x \end{matrix} ∣ ∣ ∣ ∣ ∣

then the constant term in the expansion is

2 (^{n} C_{0}) + \frac{3}{2} (^{n} C_{1}) + \frac{4}{3} (^{n} C_{2}) + \frac{5}{4} (^{n} C_{3}) . . .

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