Problem 8
Question
If \(\sum_{n=1}^{\infty} x^{n-1}=a\) and \(\sum_{n=1}^{\infty} y^{n-1}=b\) where \(|x|,|y|<1\), then \(\sum_{n=1}^{\infty}(x y)^{n-1}=\) (A) \(a b\) (B) \(\frac{a+b-1}{a b}\) (C) \(\frac{1}{1-a b}\) (D) \(\frac{a b}{a+b-1}\)
Step-by-Step Solution
Verified Answer
(C) \(\frac{1}{1-ab}\)
1Step 1: Identify the Given Series
The series given are \( \sum_{n=1}^{\infty} x^{n-1}=a \) and \( \sum_{n=1}^{\infty} y^{n-1}=b \). These are geometric series with first term \( 1 \) and common ratios \( x \) and \( y \), respectively.
2Step 2: Recall the Formula for a Geometric Series
The sum of an infinite geometric series \( \sum_{n=0}^{\infty} r^{n} \) is \( \frac{1}{1-r} \), provided that \( |r|<1 \). The given series \( \sum_{n=1}^{\infty} x^{n-1} \) can be rewritten as \( \sum_{n=0}^{\infty} x^{n} \) which simplifies to \( \frac{1}{1-x} \). Similarly, for \( y \), it is \( \frac{1}{1-y} \).
Key Concepts
Infinite SeriesCommon RatioConvergence of Series
Infinite Series
An infinite series is a sum of infinitely many terms. It continues without end or limit.
In mathematics, we often deal with infinite series to understand patterns or solve complex problems.
In the exercise, we encounter infinite series of the form \[ \sum_{n=1}^{\infty} x^{n-1} = a \] and \[ \sum_{n=1}^{\infty} y^{n-1} = b \].
These represent geometric series that include infinite terms starting from the power of zero.
The infinite nature of such series makes them valuable for precise calculations in various fields.
In mathematics, we often deal with infinite series to understand patterns or solve complex problems.
In the exercise, we encounter infinite series of the form \[ \sum_{n=1}^{\infty} x^{n-1} = a \] and \[ \sum_{n=1}^{\infty} y^{n-1} = b \].
These represent geometric series that include infinite terms starting from the power of zero.
The infinite nature of such series makes them valuable for precise calculations in various fields.
Common Ratio
The concept of a common ratio is key to understanding geometric series.
In any geometric series, the common ratio is the constant factor between consecutive terms.
If the terms of a series are: 1, 3, 9, 27, ..., the common ratio is 3 because each term is 3 times the previous one.For our given series:
In any geometric series, the common ratio is the constant factor between consecutive terms.
If the terms of a series are: 1, 3, 9, 27, ..., the common ratio is 3 because each term is 3 times the previous one.For our given series:
- The common ratio for \(\sum_{n=1}^{\infty} x^{n-1}=a\) is \x\.
- The common ratio for \(\sum_{n=1}^{\infty} y^{n-1}=b\) is \y\.
Convergence of Series
Convergence of a series refers to whether the series approaches a finite value as more terms are added.
A series converges if adding more terms results in a sum that stabilizes to a particular number.In the case of geometric series, the series will converge if the absolute value of the common ratio is less than one.
Mathematically, a geometric series \(\sum_{n=0}^{\infty} r^{n}\) converges to \(\frac{1}{1-r}\) if \(|r| < 1\).
A series converges if adding more terms results in a sum that stabilizes to a particular number.In the case of geometric series, the series will converge if the absolute value of the common ratio is less than one.
Mathematically, a geometric series \(\sum_{n=0}^{\infty} r^{n}\) converges to \(\frac{1}{1-r}\) if \(|r| < 1\).
- In our exercise, both \(|x| < 1\) and \(|y| < 1\), ensuring that both series are convergent.
- This property is crucial, as it allows us to represent and manipulate these infinite sums neatly as \(\frac{1}{1-x}\) and \(\frac{1}{1-y}\), respectively.
Other exercises in this chapter
Problem 6
Let \(S_{n}(1 \leq n \leq 9)\) denotes the sum of \(n\) terms of series \(1+22+333+\ldots+999999999\), then for \(2 \leq n \leq 9\) (A) \(S_{n}-S_{n-1}=\frac{1}
View solution Problem 7
If \(\log _{\sqrt{5}} x+\log _{5^{n}} x+\log _{5^{4}} x+\ldots\) upto 7 terms \(=35\), then \(x\) is equal to (A) 5 (B) 25 (C) 125 (D) None of these
View solution Problem 9
Let \(p, q, r \in R^{+}\)and \(27 p q r \geq(p+q+r)^{3}\) and \(3 p+4 q\) \(+5 r=12\) then \(p^{3}+q^{4}+r^{5}\) is equal to (A) 3 (B) 6 (C) 2 (D) None of these
View solution Problem 10
The sum of the series \(\frac{1}{1+1^{2}+1^{4}}+\frac{2}{1+2^{2}+2^{4}}+\frac{3}{1+3^{2}+3^{4}}+\ldots\) to \(n\) terms is (A) \(\frac{n\left(n^{2}+1\right)}{n^
View solution