Problem 101
Question
If \(|a|<1\) and \(|b|<1\), then the sum of the series \(1+(1+a) b+\left(1+a+a^{2}\right) b^{2}+\left(1+a+a^{2}+a^{3}\right) b^{3}+\) \(\ldots \infty\) is equal to (A) \(\frac{1}{(1-b)(1-a b)}\) (B) \(\frac{1}{(1-a)(1-a b)}\) (C) \(\frac{1}{(1-a)(1-b)}\) (D) None of these
Step-by-Step Solution
Verified Answer
The sum of the series is \(\frac{1}{(1-a)(1-b)}\); answer is (C).
1Step 1: Examine the Series
The given series is \(1 + (1+a)b + (1+a+a^2)b^2 + (1+a+a^2+a^3)b^3 + \ldots \). Notice that each coefficient of \(b^n\) is a geometric series in terms of \(a\).
2Step 2: Identify the Pattern in Coefficients
Each coefficient before \(b^n\) follows the pattern \(1 + a + a^2 + \ldots + a^{n}\). This is the sum of a geometric series with ratio \(a\) and \(n+1\) terms.
3Step 3: Simplify the Geometric Series for Coefficients
For a geometric series \(1 + a + a^2 + \ldots + a^{n}\), the formula for the sum is \(\frac{1-a^{n+1}}{1-a}\). When \(|a| < 1\), as \(n\) approaches infinity, \(a^{n+1} \to 0\), giving us \(\frac{1}{1-a}\) as the sum.
4Step 4: Rewrite the Series Using Simplified Coefficients
Substitute the sum of the coefficients into the series: \(\frac{1}{1-a} + \frac{1}{1-a}b + \frac{1}{1-a}b^2 + \ldots \). This results in the series \(\frac{1}{1-a}(1+b+b^2+\ldots)\).
5Step 5: Sum the Series in Terms of \(b\)
The series \(\sum b^n\) is geometric with a common ratio of \(b\), which can be summed to \(\frac{1}{1-b}\) given \(|b| < 1\). So the complete sum is \(\frac{1}{1-a} \cdot \frac{1}{1-b}\).
6Step 6: Compare with the Given Options
The result from the previous step \(\frac{1}{(1-a)(1-b)}\) corresponds to option (C) in the problem.
Key Concepts
Sum of Infinite SeriesGeometric ProgressionConvergence of Series
Sum of Infinite Series
The sum of an infinite series is a fundamental concept in mathematics. An infinite series is essentially an ongoing sum that continues forever. In simple terms, if you have an endless chain of terms, you add them up to see what the total comes to be. While it sounds like these sums should be infinite themselves, that's not always the case. Some infinite series actually converge, which means they approach a specific value as more terms are added.
Imagine stacking numbers one after another, and despite adding more, they settle toward one particular sum. It's like pouring sand into a jar; you might keep doing it, but after a while, the amount doesn't change much. To find the sum of these types of series, we use specific formulas depending on the pattern of the series.
Imagine stacking numbers one after another, and despite adding more, they settle toward one particular sum. It's like pouring sand into a jar; you might keep doing it, but after a while, the amount doesn't change much. To find the sum of these types of series, we use specific formulas depending on the pattern of the series.
- One common scenario is the geometric series, which we'll explore more in the next section.
- The sum of an infinite geometric series converges to a specific value if certain criteria are met.
Geometric Progression
A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. Let's break it down:
If you start a series with 1 and then multiply each subsequent term by, say, 3, the series would be 1, 3, 9, 27, and so forth.
Mathematically, a geometric sequence can be represented as: \( a, ar, ar^2, ar^3, \ldots \) where \( a \) is the first term and \( r \) is the common ratio.
This pattern forms the backbone of many problems involving series and allows us to determine several properties of the sequence:
If you start a series with 1 and then multiply each subsequent term by, say, 3, the series would be 1, 3, 9, 27, and so forth.
Mathematically, a geometric sequence can be represented as: \( a, ar, ar^2, ar^3, \ldots \) where \( a \) is the first term and \( r \) is the common ratio.
This pattern forms the backbone of many problems involving series and allows us to determine several properties of the sequence:
- If \(|r| < 1\), the series converges, meaning it approaches a finite sum.
- Conversely, if \(|r| \geq 1\), the series diverges, making it grow indefinitely or oscillate.
Convergence of Series
Convergence of a series refers to the behavior of the sum of its terms. Specifically, a series converges if as more terms are added, it approaches a specific value rather than shooting off to infinity or wiggling between different numbers.
A good rule of thumb for determining convergence in series is examining the terms and their behavior.
For example, the series \(1 + b + b^2 + b^3 + \ldots\) converges if the absolute value of the common ratio \(|b| < 1\).
A good rule of thumb for determining convergence in series is examining the terms and their behavior.
For example, the series \(1 + b + b^2 + b^3 + \ldots\) converges if the absolute value of the common ratio \(|b| < 1\).
- If aside from any large initial upsets, the series flattens out to a stable sum, we say it converges, like water flowing into a pool and reaching a steady level.
- This is particularly important in infinite geometric series, where each successive term is a fraction of the last, getting smaller and smaller.
Other exercises in this chapter
Problem 98
If \(S_{1}, S_{2}\) and \(S_{3}\) denote the sums up to \(n>1\) terms of three sequences in A.P. whose first terms are unity and common differences are in H.P.
View solution Problem 100
Let \(a\) be a fixed real number such that \(\frac{a-x}{p x}=\frac{a-y}{q y}=\frac{a-z}{r z}\)If \(p, q, \mathrm{r}\) are in A.P. then \(x, y, z\) are in (A) A.
View solution Problem 103
The sixth term of an A.P. is equal to \(2 .\) The value of the common difference of the A.P. which makes the product \(a_{1} a_{4} a_{5}\) greatest, is (A) \(\f
View solution Problem 106
If the first and the \((2 n-1)\) th terms of an A.P., G.P. and H.P. are equal and their \(n\)th terms are \(a, b, c\) respectively, then (A) \(a=b=c\) (B) \(a \
View solution