1-4 - 1-42 - 1-43 - - - - - 1-4n - 1

Given series is in GP with common ratio 14 ⇒ sum of terms in GP =a(r^n 1)r 1 ⇒14 ( (14) ^n 1 1)14 1 =13 (1 (14) ^n 1) ...

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Sum of Coefficients of All Terms

1/4 + 1/42 + ...

Question

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

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Solution

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4 C_{0} + \frac{4^{2}}{2} . C_{1} + \frac{4^{3}}{3} C_{2} + . . . . . . . . . . . . + \frac{4^{n + 1}}{n + 1} C_{n} = \frac{5^{n + 1} - 1}{n + 1}

\frac{4^{3} - 1}{3} = 4^{2} + 4 + 1

1 + (\frac{1}{2} + \frac{1}{3}) \frac{1}{4} + (\frac{1}{4} + \frac{1}{5}) \frac{1}{4^{2}} + (\frac{1}{6} + \frac{1}{7}) \frac{1}{4^{3}} + . . . \infty = {log}_{e} \sqrt{b}

\frac{b}{4}

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

b_{n} = 1 - a_{n}

b_{n} > a_{n} \forall n > n_{0},

1 + \frac{1}{1!} . \frac{1}{4} + \frac{1.3}{2!} . {(\frac{1}{4})}^{2} + \frac{1.3.5}{3!} . {(\frac{1}{4})}^{3} + . . . .

\infty

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