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Question: The sum of infinite terms of the geometric progression \[\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}},\dfrac...

The sum of infinite terms of the geometric progression 2+12−1,12−2,12,.....\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}},\dfrac{1}{{2 - \sqrt 2 }},\dfrac{1}{2},..... is
A. 2(2+1)2\sqrt 2 {\left( {\sqrt 2 + 1} \right)^2}
B. (2+1)2{\left( {\sqrt 2 + 1} \right)^2}
C. 525\sqrt 2
D. 32+53\sqrt 2 + \sqrt 5

Explanation

Solution

Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio. This progression is also known as a geometric sequence of numbers that follow a pattern a,ar,ar2,.....arn−1a,ar,a{r^2},.....a{r^{n - 1}}and so on. Hence, to find the sum of infinite terms of the geometric progression, we need to apply the formula, S∞=a1−r{S_\infty } = \dfrac{a}{{1 - r}}and simplify the terms.

Formula used:
S∞=a1−r{S_\infty } = \dfrac{a}{{1 - r}}
S∞{S_\infty } = Sum of infinite geometric progression
a = First term
r = Common ratio
n = Number of terms

Complete step by step solution:
Let us write the given geometric progression sequence as:
2+12−1,12−2,12,.....\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}},\dfrac{1}{{2 - \sqrt 2 }},\dfrac{1}{2},..... which is represented as:(a1,a2,a3,.....)\left( {{a_1},{a_2},{a_3},.....} \right)
Here,
a is the first term of the sequence i.e.,a=2+12−1a = \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}
r is the common ratio i.e., r=a2a1r = \dfrac{{{a_2}}}{{{a_1}}}
Substitute the value of a2{a_2} and a1{a_1} as:
=12−22+12−1= \dfrac{{\dfrac{1}{{2 - \sqrt 2 }}}}{{\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}}}
Evaluating the terms, we have:
=12−2×2−12+1= \dfrac{1}{{2 - \sqrt 2 }} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}
Hence, the common ratio of the series is:
r=12(2+1)r = \dfrac{1}{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}
The formula to find the sum to infinity of the given GP is:
S∞=a1−r;−1<r<1{S_\infty } = \dfrac{a}{{1 - r}}; - 1 < r < 1
Therefore, the sum is:
S∞=a1−r{S_\infty } = \dfrac{a}{{1 - r}}
Substitute the value of a and r as:
=2+12−11−12(2+1)= \dfrac{{\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}}}{{1 - \dfrac{1}{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}}}
Evaluating the terms, we get:
=2+12−1×2(2+1)(1+2)= \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}} \times \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{\left( {1 + \sqrt 2 } \right)}}
Simplifying the numerator and denominator common terms, we get:
=2(2+1)2= \sqrt 2 {\left( {\sqrt 2 + 1} \right)^2}

So, the correct answer is Option A.

Note: We must note that, when we divide any succeeding term from its preceding term, then we get the value equal to common ratio. And if three quantities are in GP, then the middle one is called the geometric mean of the other two terms. If we are asked to find the sum of n terms of geometric series then, we have the formula as: Sn=a(rn−1)r−1;r≠1{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}};r \ne 1.