Find the general term: $1 + 5 + 19 + 65 + . . . . . .$ to n terms

3^{n - 1}

2^{n}

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3^{n}

2^{n}

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3^{n + 1}

2^{n}

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3^{n - 1}

2^{n + 1}

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Solution

The correct option is B $3^{n}$ - $2^{n}$
$1 = 3^{1} - 2^{1} 5 = 3^{2} - 2^{2} 19 = 3^{3} - 2^{3} ∴ n^{t h} t e r m = 3^{n} - 2^{n}$

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Similar questions

The (n+1)th term from the end in the expansion of $(2 x - \frac{1}{x})^{3 n}$ is:

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

{lim}_{n \to \infty} (\frac{2^{n + 1} + 3^{n + 1}}{2^{n} + 3^{n}})

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

(1 - \frac{1}{n}) + (1 - \frac{2}{n}) + (1 - \frac{3}{n}) + . . . upto n terms

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