Question

If the sum of the series $(\frac{1}{2}-\frac{1}{3})+(\frac{1}{2^2}-\frac{1}{2.3}+\frac{1}{3^2})+(\frac{1}{2^3}-\frac{1}{2^2.3}+\frac{1}{2.3^2}-\frac{1}{3^3})+(\frac{1}{2^4}-\frac{1}{2^3.3}+\frac{1}{2^2.3^2}-\frac{1}{2.3^3}+\frac{1}{3^4})+....$ is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha + 3\beta$ is equal to....

Answer 1

7

Answer 2

Let the given series be denoted by $S$. The series is $S = (\frac{1}{2}-\frac{1}{3})+(\frac{1}{2^2}-\frac{1}{2.3}+\frac{1}{3^2})+(\frac{1}{2^3}-\frac{1}{2^2.3}+\frac{1}{2.3^2}-\frac{1}{3^3})+(\frac{1}{2^4}-\frac{1}{2^3.3}+\frac{1}{2^2.3^2}-\frac{1}{2.3^3}+\frac{1}{3^4})+....$

Let $a = \frac{1}{2}$ and $b = \frac{1}{3}$. The terms in the series are: First term: $T_1 = a - b$ Second term: $T_2 = a^2 - ab + b^2$ Third term: $T_3 = a^3 - a^2b + ab^2 - b^3$ Fourth term: $T_4 = a^4 - a^3b + a^2b^2 - ab^3 + b^4$ The $n$-th term of the series is $T_n = a^n - a^{n-1}b + a^{n-2}b^2 - \dots + (-1)^n b^n$.

This is a geometric series with first term $a^n$, common ratio $-b/a$, and $n+1$ terms. The sum of this geometric series is given by the formula $\text{Sum} = \text{First term} \times \frac{1 - (\text{common ratio})^{\text{number of terms}}}{1 - \text{common ratio}}$. $T_n = a^n \frac{1 - (-b/a)^{n+1}}{1 - (-b/a)} = a^n \frac{1 - (-1)^{n+1} (b/a)^{n+1}}{1 + b/a} = a^n \frac{1 - (-1)^{n+1} b^{n+1}/a^{n+1}}{(a+b)/a}$ $T_n = \frac{a^{n+1}}{a+b} (1 - (-1)^{n+1} b^{n+1}/a^{n+1}) = \frac{a^{n+1} - (-1)^{n+1} b^{n+1}}{a+b}$.

We need to find the sum of the series $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{a^{n+1} - (-1)^{n+1} b^{n+1}}{a+b}$. $S = \frac{1}{a+b} \sum_{n=1}^{\infty} (a^{n+1} - (-1)^{n+1} b^{n+1})$. $S = \frac{1}{a+b} \left( \sum_{n=1}^{\infty} a^{n+1} - \sum_{n=1}^{\infty} (-1)^{n+1} b^{n+1} \right)$.

Let's evaluate the first infinite sum $\sum_{n=1}^{\infty} a^{n+1}$. This is $a^2 + a^3 + a^4 + \dots$. This is an infinite geometric series with first term $A = a^2$ and common ratio $r = a$. Since $a = \frac{1}{2}$, $|a| < 1$, the sum converges to $\frac{A}{1-r} = \frac{a^2}{1-a}$. $\sum_{n=1}^{\infty} a^{n+1} = \frac{(1/2)^2}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{4} \times 2 = \frac{1}{2}$.

Let's evaluate the second infinite sum $\sum_{n=1}^{\infty} (-1)^{n+1} b^{n+1}$. This is $(-1)^2 b^2 + (-1)^3 b^3 + (-1)^4 b^4 + \dots = b^2 - b^3 + b^4 - \dots$. This is an infinite geometric series with first term $A = b^2$ and common ratio $r = -b$. Since $b = \frac{1}{3}$, $|-b| = |-1/3| < 1$, the sum converges to $\frac{A}{1-r} = \frac{b^2}{1 - (-b)} = \frac{b^2}{1+b}$. $\sum_{n=1}^{\infty} (-1)^{n+1} b^{n+1} = \frac{(1/3)^2}{1 + 1/3} = \frac{1/9}{4/3} = \frac{1}{9} \times \frac{3}{4} = \frac{3}{36} = \frac{1}{12}$.

Now substitute these values back into the expression for $S$. $a+b = \frac{1}{2} + \frac{1}{3} = \frac{3+2}{6} = \frac{5}{6}$. $S = \frac{1}{5/6} \left( \frac{1}{2} - \frac{1}{12} \right)$. $S = \frac{6}{5} \left( \frac{6}{12} - \frac{1}{12} \right) = \frac{6}{5} \left( \frac{5}{12} \right) = \frac{6 \times 5}{5 \times 12} = \frac{30}{60} = \frac{1}{2}$.

The sum of the series is $S = \frac{1}{2}$. We are given that the sum is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime. Comparing $S = \frac{1}{2}$ with $\frac{\alpha}{\beta}$, we have $\alpha = 1$ and $\beta = 2$. $\alpha = 1$ and $\beta = 2$ are co-prime (their greatest common divisor is 1).

We need to find the value of $\alpha + 3\beta$. $\alpha + 3\beta = 1 + 3(2) = 1 + 6 = 7$.