OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Exercise 14 (H)

Question 1. \( 1 + 2x + 3x^2 + 4x^3 + ......\)
Answer: Let \( S_n = 1 + 2x + 3x^2 + 4x^3 + ......\) up to n terms.
This sequence is an arithmetico-geometric series, which combines properties of both arithmetic and geometric progressions. We can write it as \( S_n = 1 + 2x + 3x^2 + 4x^3 + \ldots + (n-1) x^{n-2}+n x^{n-1} \).
In simple words: This is a special type of series where each term is a product of terms from an arithmetic progression and a geometric progression.

🎯 Exam Tip: When dealing with arithmetico-geometric series, always multiply the series by the common ratio of the geometric part and subtract the new series from the original to simplify it.

Question 2. \( 1 + 3x +5x^2 + 7x^3 + ......\)
Answer: Let \( S_n = 1 + 3x + 5x^2 + 7x^3 + \ldots + (2n - 3) x^{n-2} + (2n - 1) x^{n-1} \) (1)
This is an arithmetico-geometric progression (A.G.P.).
Multiply equation (1) by \( x \):
\( xS_n = x + 3x^2 + 5x^3 + \ldots + (2n - 3) x^{n-1} + (2n - 1) x^n \) (2)
Subtract equation (2) from equation (1):
\( (1 - x)S_n = 1 + 2x + 2x^2 + 2x^3 - \ldots + 2x^{n-1} - (2n - 1) x^n \)
\( \implies (1 - x)S_n = 1 + 2x(1 + x + x^2 + \ldots + x^{n-2}) - (2n - 1)x^n \)
The part \( (1 + x + x^2 + \ldots + x^{n-2}) \) is a geometric series sum.
\( \implies (1 - x)S_n = 1 + 2x \left[ \frac{1-x^{n-1}}{1-x} \right] - (2n - 1)x^n \)
\( \implies (1 - x)S_n = 1 + \frac{2x}{1-x} - \frac{2x^n}{1-x} - (2n - 1)x^n \)
\( \implies S_n = \frac{1}{1-x} + \frac{2x}{(1-x)^2} - \frac{2x^n}{(1-x)^2} - \frac{(2n-1)x^n}{1-x} \)
\( \implies S_n = \frac{1-x+2x}{(1-x)^2} - \frac{2x^n + (2n-1)x^n(1-x)}{(1-x)^2} \)
\( \implies S_n = \frac{1+x}{(1-x)^2} - \frac{2x^n + (2n-1)x^n - (2n-1)x^{n+1}}{(1-x)^2} \)
\( \implies S_n = \frac{1+x}{(1-x)^2} - \frac{(2n+1)x^n - (2n-1)x^{n+1}}{(1-x)^2} \)
In simple words: To find the sum, we subtract the series multiplied by \( x \) from the original series. This helps to make a new series that is easier to sum, often a geometric progression.

🎯 Exam Tip: Remember to carefully manage the exponents and signs when subtracting the two series. Mistakes often happen with the last few terms.

Question 3. \( 2.1 + 3.2 + 4.4 + 5.8 + ......\)
Answer: Let \( S_n = 2 \cdot 1 + 3 \cdot 2 + 4 \cdot 4 + 5 \cdot 8 + \ldots + (n+1) 2^{n-1} \) (1)
Multiply equation (1) by the common ratio of the geometric part, which is 2:
\( 2S_n = 2 \cdot 2 + 3 \cdot 4 + 4 \cdot 8 + \ldots + n 2^{n-1} + (n+1) 2^n \) (2)
Subtract equation (2) from equation (1):
\( -S_n = 2 \cdot 1 + (3 \cdot 2 - 2 \cdot 2) + (4 \cdot 4 - 3 \cdot 4) + (5 \cdot 8 - 4 \cdot 8) + \ldots + ((n+1)2^{n-1} - n2^{n-1}) - (n+1)2^n \)
\( -S_n = 2 + 1 \cdot 2 + 1 \cdot 4 + 1 \cdot 8 + \ldots + 1 \cdot 2^{n-1} - (n+1)2^n \)
\( -S_n = 2 + 2 + 4 + 8 + \ldots + 2^{n-1} - (n+1)2^n \)
The terms \( 2 + 4 + \ldots + 2^{n-1} \) form a geometric series with first term \( a=2 \), common ratio \( r=2 \), and \( n-1 \) terms.
The sum of this geometric series is \( \frac{a(r^{number\ of\ terms} - 1)}{r-1} = \frac{2(2^{n-1} - 1)}{2-1} = 2(2^{n-1} - 1) = 2^n - 2 \).
So, \( -S_n = 2 + (2^n - 2) - (n+1)2^n \)
\( -S_n = 2^n - (n+1)2^n \)
\( -S_n = 2^n - n2^n - 2^n \)
\( -S_n = -n2^n \)
\( S_n = n2^n \)
In simple words: This problem involves a series where each term is a product of two numbers. By multiplying the original series by 2 and then subtracting, we get a simpler series that we can sum up, leading to the final result of \( n \cdot 2^n \).

🎯 Exam Tip: When the coefficients form an A.P. and the other factor forms a G.P., the method of multiplying by the common ratio of the G.P. and subtracting is the most effective way to find the sum.

Question 4. \( \frac{1}{2} + \frac{3}{6} + \frac{5}{18} + ......\)
Answer: Let \( S = \frac{1}{2} + \frac{3}{6} + \frac{5}{18} + \ldots \infty \) (1)
This is an arithmetico-geometric series. The common ratio of the geometric part is \( r = \frac{1}{3} \).
Multiply equation (1) by \( \frac{1}{3} \):
\( \frac{1}{3}S = \frac{1}{6} + \frac{3}{18} + \frac{5}{54} + \ldots \infty \) (2)
Subtract equation (2) from equation (1):
\( S - \frac{1}{3}S = \frac{1}{2} + (\frac{3}{6} - \frac{1}{6}) + (\frac{5}{18} - \frac{3}{18}) + \ldots \infty \)
\( \frac{2}{3}S = \frac{1}{2} + \frac{2}{6} + \frac{2}{18} + \ldots \infty \)
\( \frac{2}{3}S = \frac{1}{2} + \frac{1}{3} + \frac{1}{9} + \ldots \infty \)
The terms \( \frac{1}{3} + \frac{1}{9} + \ldots \infty \) form an infinite geometric series with first term \( a = \frac{1}{3} \) and common ratio \( r = \frac{1}{3} \).
The sum of an infinite geometric series is \( S_\infty = \frac{a}{1-r} \).
So, \( \frac{1}{3} + \frac{1}{9} + \ldots \infty = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \).
Therefore, \( \frac{2}{3}S = \frac{1}{2} + \frac{1}{2} \)
\( \frac{2}{3}S = 1 \)
\( S = \frac{3}{2} \)
In simple words: We find the sum of this special type of series by subtracting a shifted version of itself. This turns most of the series into a simpler geometric progression, which we can then easily sum up to get 3/2.

🎯 Exam Tip: For infinite arithmetico-geometric series, ensure the absolute value of the common ratio of the geometric progression is less than 1 (i.e., \( |r| < 1 \)) for the sum to converge.

Question 5. \( \frac{3}{2} – \frac{5}{6} + \frac{7}{18} – ......\)
Answer: Let \( S = \frac{3}{2} – \frac{5}{6} + \frac{7}{18} - \ldots \infty \) (1)
This is an arithmetico-geometric series. The common ratio of the geometric part is \( r = -\frac{1}{3} \).
Multiply equation (1) by \( -\frac{1}{3} \):
\( -\frac{1}{3}S = -\frac{3}{6} + \frac{5}{18} - \frac{7}{54} + \ldots \infty \) (2)
Subtract equation (2) from equation (1):
\( S - (-\frac{1}{3}S) = \frac{3}{2} - (-\frac{3}{6}) + (-\frac{5}{6} - \frac{5}{18}) + (\frac{7}{18} - (-\frac{7}{54})) + \ldots \infty \)
\( \frac{4}{3}S = \frac{3}{2} + \frac{3}{6} - \frac{5}{6} - \frac{5}{18} + \frac{7}{18} + \frac{7}{54} - \ldots \infty \)
\( \frac{4}{3}S = \frac{3}{2} + (-\frac{5}{6} - (-\frac{3}{6})) + (\frac{7}{18} - (-\frac{5}{18})) + \ldots \infty \)
\( \frac{4}{3}S = \frac{3}{2} + (\frac{-5+3}{6}) + (\frac{7+5}{18}) + \ldots \infty \)
\( \frac{4}{3}S = \frac{3}{2} - \frac{2}{6} + \frac{2}{18} - \frac{2}{54} + \ldots \infty \)
\( \frac{4}{3}S = \frac{3}{2} - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots \infty \)
The terms \( -\frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots \infty \) form an infinite geometric series with first term \( a = -\frac{1}{3} \) and common ratio \( r = -\frac{1}{3} \).
The sum of this infinite geometric series is \( S_\infty = \frac{a}{1-r} = \frac{-\frac{1}{3}}{1-(-\frac{1}{3})} = \frac{-\frac{1}{3}}{1+\frac{1}{3}} = \frac{-\frac{1}{3}}{\frac{4}{3}} = -\frac{1}{4} \).
So, \( \frac{4}{3}S = \frac{3}{2} + (-\frac{1}{4}) \)
\( \frac{4}{3}S = \frac{6-1}{4} = \frac{5}{4} \)
\( S = \frac{5}{4} \times \frac{3}{4} \)
\( S = \frac{15}{16} \)
In simple words: This problem involves a series that alternates signs and combines arithmetic and geometric patterns. We solve it by multiplying the series by the common ratio, subtracting it from the original, and then summing the resulting geometric progression.

🎯 Exam Tip: Be very careful with the signs when the common ratio is negative. Each term in the new series will also have its sign changed, which can lead to errors if not handled properly.

Question 6. \( 1 - \frac{2}{5} + \frac{3}{5^2} - \frac{4}{5^3} + ......\)
Answer: Let \( S = 1 - \frac{2}{5} + \frac{3}{5^2} - \frac{4}{5^3} + \ldots \infty \) (1)
This is an arithmetico-geometric series. The common ratio of the geometric part is \( r = -\frac{1}{5} \).
Multiply equation (1) by \( -\frac{1}{5} \):
\( -\frac{1}{5}S = -\frac{1}{5} + \frac{2}{5^2} - \frac{3}{5^3} + \ldots \infty \) (2)
Subtract equation (2) from equation (1):
\( S - (-\frac{1}{5}S) = 1 + (-\frac{2}{5} - (-\frac{1}{5})) + (\frac{3}{5^2} - \frac{2}{5^2}) + \ldots \infty \)
\( \frac{6}{5}S = 1 - \frac{1}{5} + \frac{1}{5^2} - \frac{1}{5^3} + \ldots \infty \)
The right-hand side is an infinite geometric series with first term \( a = 1 \) and common ratio \( r = -\frac{1}{5} \).
The sum of an infinite geometric series is \( S_\infty = \frac{a}{1-r} \).
So, \( 1 - \frac{1}{5} + \frac{1}{5^2} - \ldots \infty = \frac{1}{1-(-\frac{1}{5})} = \frac{1}{1+\frac{1}{5}} = \frac{1}{\frac{6}{5}} = \frac{5}{6} \).
Therefore, \( \frac{6}{5}S = \frac{5}{6} \)
\( S = \frac{5}{6} \times \frac{5}{6} \)
\( S = \frac{25}{36} \)
In simple words: We find the sum of this alternating series to infinity by adding a scaled version of the series to itself. This changes it into a simple geometric series, which we then sum to get the final answer.

🎯 Exam Tip: When the terms alternate in sign, the common ratio \( r \) will be negative. Make sure to use \( 1-r \) in the denominator for the sum of an infinite G.P. formula.

Question 7. Sum up the series \( \frac{2}{3} + \frac{5}{9} + \frac{8}{27} + \frac{11}{81} + ......\) to n terms and hence find the sum to infinity.
Answer: Let \( S_n = \frac{2}{3} + \frac{5}{9} + \frac{8}{27} + \frac{11}{81} + \ldots + \frac{3n-1}{3^n} \) (1)
This is an arithmetico-geometric series. The common ratio of the geometric part is \( r = \frac{1}{3} \).
Multiply equation (1) by \( \frac{1}{3} \):
\( \frac{1}{3}S_n = \frac{2}{9} + \frac{5}{27} + \frac{8}{81} + \ldots + \frac{3n-4}{3^n} + \frac{3n-1}{3^{n+1}} \) (2)
Subtract equation (2) from equation (1):
\( S_n - \frac{1}{3}S_n = \frac{2}{3} + (\frac{5}{9} - \frac{2}{9}) + (\frac{8}{27} - \frac{5}{27}) + (\frac{11}{81} - \frac{8}{81}) + \ldots + (\frac{3n-1}{3^n} - \frac{3n-4}{3^n}) - \frac{3n-1}{3^{n+1}} \)
\( \frac{2}{3}S_n = \frac{2}{3} + \frac{3}{9} + \frac{3}{27} + \frac{3}{81} + \ldots + \frac{3}{3^n} - \frac{3n-1}{3^{n+1}} \)
\( \frac{2}{3}S_n = \frac{2}{3} + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^{n-1}} - \frac{3n-1}{3^{n+1}} \)
The terms \( \frac{1}{3} + \frac{1}{9} + \ldots + \frac{1}{3^{n-1}} \) form a geometric series with first term \( a = \frac{1}{3} \), common ratio \( r = \frac{1}{3} \), and \( n-1 \) terms.
The sum of this geometric series is \( \frac{a(1-r^{number\ of\ terms})}{1-r} = \frac{\frac{1}{3}(1-(\frac{1}{3})^{n-1})}{1-\frac{1}{3}} = \frac{\frac{1}{3}(1-\frac{1}{3^{n-1}})}{\frac{2}{3}} = \frac{1}{2}(1-\frac{1}{3^{n-1}}) = \frac{1}{2} - \frac{1}{2 \cdot 3^{n-1}} \).
So, \( \frac{2}{3}S_n = \frac{2}{3} + (\frac{1}{2} - \frac{1}{2 \cdot 3^{n-1}}) - \frac{3n-1}{3^{n+1}} \)
\( \frac{2}{3}S_n = \frac{4+3}{6} - \frac{1}{2 \cdot 3^{n-1}} - \frac{3n-1}{3^{n+1}} \)
\( \frac{2}{3}S_n = \frac{7}{6} - \frac{3}{2 \cdot 3^n} - \frac{3n-1}{3^{n+1}} \)
\( S_n = \frac{3}{2} \left( \frac{7}{6} - \frac{3}{2 \cdot 3^n} - \frac{3n-1}{3^{n+1}} \right) \)
\( S_n = \frac{7}{4} - \frac{9}{4 \cdot 3^n} - \frac{3(3n-1)}{2 \cdot 3^{n+1}} \)
\( S_n = \frac{7}{4} - \frac{1}{4 \cdot 3^{n-2}} - \frac{3n-1}{2 \cdot 3^n} \)
\( S_n = \frac{7}{4} - \frac{9}{4 \cdot 3^n} - \frac{3n-1}{2 \cdot 3^n} = \frac{7}{4} - \frac{9+2(3n-1)}{4 \cdot 3^n} = \frac{7}{4} - \frac{9+6n-2}{4 \cdot 3^n} = \frac{7}{4} - \frac{6n+7}{4 \cdot 3^n} \).

To find the sum to infinity, \( S_\infty \), we let \( n \rightarrow \infty \).
As \( n \rightarrow \infty \), \( \frac{6n+7}{4 \cdot 3^n} \rightarrow 0 \). (Since exponential growth \( 3^n \) is much faster than linear growth \( 6n+7 \)).
So, \( S_\infty = \frac{7}{4} - 0 \)
\( S_\infty = \frac{7}{4} \)
In simple words: First, we find the sum of the series up to 'n' terms by using the subtraction method for arithmetico-geometric series. Then, to find the sum to infinity, we look at what happens to the terms as 'n' becomes very, very large. In this case, most parts become tiny and disappear, leaving us with a simple fraction.

🎯 Exam Tip: When finding the sum to infinity for an A.G.P., remember that terms involving \( n \) multiplied by a power of \( r \) (where \( |r| < 1 \)) will approach zero as \( n \to \infty \).

Question 8. \( 1 + 4x^2 + 7x^4 + ......\)
Answer: Let \( S = 1 + 4x^2 + 7x^4 + \ldots \infty \) (1)
This is an arithmetico-geometric series where the arithmetic part is \( 1, 4, 7, \ldots \) (with common difference 3) and the geometric part is \( 1, x^2, x^4, \ldots \) (with common ratio \( x^2 \)).
Multiply equation (1) by \( x^2 \):
\( x^2S = x^2 + 4x^4 + 7x^6 + \ldots \infty \) (2)
Subtract equation (2) from equation (1):
\( S - x^2S = 1 + (4x^2 - x^2) + (7x^4 - 4x^4) + \ldots \infty \)
\( (1 - x^2)S = 1 + 3x^2 + 3x^4 + 3x^6 + \ldots \infty \)
\( (1 - x^2)S = 1 + 3x^2(1 + x^2 + x^4 + \ldots \infty) \)
The terms \( 1 + x^2 + x^4 + \ldots \infty \) form an infinite geometric series with first term \( a = 1 \) and common ratio \( r = x^2 \). For the sum to converge, we assume \( |x^2| < 1 \).
The sum of this infinite geometric series is \( \frac{a}{1-r} = \frac{1}{1-x^2} \).
So, \( (1 - x^2)S = 1 + 3x^2 \left( \frac{1}{1-x^2} \right) \)
\( (1 - x^2)S = \frac{1-x^2+3x^2}{1-x^2} \)
\( (1 - x^2)S = \frac{1+2x^2}{1-x^2} \)
\( S = \frac{1+2x^2}{(1-x^2)^2} \)
In simple words: This problem asks for the sum to infinity of a series where powers of \( x^2 \) are multiplied by terms that increase by 3 each time. By subtracting the series multiplied by \( x^2 \) from the original, we can simplify it to a basic geometric series and then find its sum.

🎯 Exam Tip: Be careful to identify the correct common ratio for the geometric part, which here is \( x^2 \), not \( x \).

Question 9. \( 1 - x + 2x^2 – 3x^3 + 4x^4 – ......\)
Answer: Let \( S = 1 - x + 2x^2 - 3x^3 + 4x^4 - \ldots \infty \) (1)
This is an arithmetico-geometric series. The common ratio of the geometric part is \( -x \).
Multiply equation (1) by \( -x \):
\( -xS = -x + x^2 - 2x^3 + 3x^4 - \ldots \infty \) (2)
Subtract equation (2) from equation (1):
\( S - (-xS) = 1 + (-x - (-x)) + (2x^2 - x^2) + (-3x^3 - (-2x^3)) + \ldots \infty \)
\( (1 + x)S = 1 + 0 + x^2 - x^3 + x^4 - \ldots \infty \)
\( (1 + x)S = 1 + x^2(1 - x + x^2 - \ldots \infty) \)
The terms \( 1 - x + x^2 - \ldots \infty \) form an infinite geometric series with first term \( a = 1 \) and common ratio \( r = -x \). For the sum to converge, we assume \( |-x| < 1 \).
The sum of this infinite geometric series is \( \frac{a}{1-r} = \frac{1}{1-(-x)} = \frac{1}{1+x} \).
So, \( (1 + x)S = 1 + x^2 \left( \frac{1}{1+x} \right) \)
\( (1 + x)S = \frac{1+x+x^2}{1+x} \)
\( S = \frac{1+x+x^2}{(1+x)^2} \)
In simple words: We are summing an alternating series to infinity where the coefficients are increasing and the powers of \( x \) also increase. By multiplying the series by \( -x \) and subtracting, we simplify it into a geometric series which can be easily summed.

🎯 Exam Tip: When the terms alternate in sign, the common ratio for the geometric part is negative. Ensure your algebra is precise, especially with the minus signs during subtraction.

Question 10. \( 1^2 + 3^2x + 5^2x^2 + 7^2x^3 + ......\)
Answer: Let \( S = 1^2 + 3^2x + 5^2x^2 + 7^2x^3 + \ldots \infty \)
\( S = 1 + 9x + 25x^2 + 49x^3 + \ldots \infty \) (1)
This is an arithmetico-geometric series. The common ratio of the geometric part is \( x \).
Multiply equation (1) by \( x \):
\( xS = x + 9x^2 + 25x^3 + 49x^4 + \ldots \infty \) (2)
Subtract equation (2) from equation (1):
\( (1 - x)S = 1 + (9x - x) + (25x^2 - 9x^2) + (49x^3 - 25x^3) + \ldots \infty \)
\( (1 - x)S = 1 + 8x + 16x^2 + 24x^3 + \ldots \infty \) (3)
The series \( 1 + 8x + 16x^2 + 24x^3 + \ldots \) is still an arithmetico-geometric progression.
Multiply equation (3) by \( x \):
\( x(1 - x)S = x + 8x^2 + 16x^3 + 24x^4 + \ldots \infty \) (4)
Subtract equation (4) from equation (3):
\( (1 - x)S - x(1 - x)S = 1 + (8x - x) + (16x^2 - 8x^2) + (24x^3 - 16x^3) + \ldots \infty \)
\( (1 - x)^2S = 1 + 7x + 8x^2 + 8x^3 + 8x^4 + \ldots \infty \)
\( (1 - x)^2S = 1 + 7x + 8x^2(1 + x + x^2 + \ldots \infty) \)
The terms \( 1 + x + x^2 + \ldots \infty \) form an infinite geometric series with first term \( a = 1 \) and common ratio \( r = x \). For the sum to converge, we assume \( |x| < 1 \).
The sum of this infinite geometric series is \( \frac{a}{1-r} = \frac{1}{1-x} \).
So, \( (1 - x)^2S = 1 + 7x + 8x^2 \left( \frac{1}{1-x} \right) \)
\( (1 - x)^2S = \frac{(1+7x)(1-x)+8x^2}{1-x} \)
\( (1 - x)^2S = \frac{1-x+7x-7x^2+8x^2}{1-x} \)
\( (1 - x)^2S = \frac{1+6x+x^2}{1-x} \)
\( S = \frac{1+6x+x^2}{(1-x)^3} \)
In simple words: This problem involves a series with squared odd numbers as coefficients. We solve it by applying the "multiply by common ratio and subtract" method twice, which simplifies the series into a basic geometric progression that can be easily summed.

🎯 Exam Tip: For series with squared or higher powers of arithmetic terms, you might need to apply the subtraction method multiple times to reduce it to a simple geometric series.

Question 11. Show that the square root of \( 3^{\frac{1}{2}} \times 9^{\frac{1}{4}} \times 27^{\frac{1}{8}} \times 81^{\frac{1}{16}} \times ......\) to infinity is 3.
Answer: Let \( S = 3^{\frac{1}{2}} \times 9^{\frac{1}{4}} \times 27^{\frac{1}{8}} \times 81^{\frac{1}{16}} \times \ldots \infty \)
We can rewrite each base as a power of 3:
\( S = 3^{\frac{1}{2}} \times (3^2)^{\frac{1}{4}} \times (3^3)^{\frac{1}{8}} \times (3^4)^{\frac{1}{16}} \times \ldots \infty \)
\( S = 3^{\frac{1}{2}} \times 3^{\frac{2}{4}} \times 3^{\frac{3}{8}} \times 3^{\frac{4}{16}} \times \ldots \infty \)
\( S = 3^{\frac{1}{2}} \times 3^{\frac{1}{2}} \times 3^{\frac{3}{8}} \times 3^{\frac{1}{4}} \times \ldots \infty \)
When multiplying powers with the same base, we add the exponents:
\( S = 3^{\left(\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \ldots \infty\right)} \)
Let \( S_1 = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \ldots \infty \) (1)
This is an arithmetico-geometric series. The common ratio of the geometric part is \( r = \frac{1}{2} \).
Multiply equation (1) by \( \frac{1}{2} \):
\( \frac{1}{2}S_1 = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} + \ldots \infty \) (2)
Subtract equation (2) from equation (1):
\( S_1 - \frac{1}{2}S_1 = \frac{1}{2} + (\frac{2}{4} - \frac{1}{4}) + (\frac{3}{8} - \frac{2}{8}) + (\frac{4}{16} - \frac{3}{16}) + \ldots \infty \)
\( \frac{1}{2}S_1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \infty \)
The right-hand side is an infinite geometric series with first term \( a = \frac{1}{2} \) and common ratio \( r = \frac{1}{2} \).
The sum of this infinite geometric series is \( S_\infty = \frac{a}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \).
So, \( \frac{1}{2}S_1 = 1 \)
\( S_1 = 2 \)
Now substitute this back into the expression for \( S \):
\( S = 3^{S_1} = 3^2 = 9 \)
The question asks for the square root of \( S \).
Square root of \( S = \sqrt{9} = 3 \).
Thus, the square root of the given product to infinity is 3. It's interesting how a complex product simplifies to a whole number.
In simple words: First, we change all the numbers into powers of 3. Then, we add all the little power numbers together. This sum of powers turns out to be a special kind of series which we can find to be 2. So, the whole big multiplication becomes \( 3^2 = 9 \). Finally, the square root of 9 is 3.

🎯 Exam Tip: Problems involving products of terms with increasing powers often simplify to a sum of exponents. Always convert all bases to a common base first, then calculate the sum of the exponents, which is often an A.G.P. or G.P.