Question

$1 /\left(3^{2}-1\right)+1 /\left(5^{2}-1\right)+1 /\left(7^{2}-1\right)+\ldots \ldots . .+1 /\left(201^{2}-1\right)$ is equal to
a. $101 / 404$
b. $101 / 408$
C. $99 / 400$
d. $25/101$

Answer 1

Hint: A sort of sequence known as geometric progression (GP) is one in which each following phrase is created by multiplying each preceding term by a fixed number, or "common ratio." This progression is sometimes referred to as a pattern-following geometric sequence of numbers. Learn development in mathematics here as well. Here, each phrase is multiplied by the common ratio to generate the subsequent term, which is a non-zero value.

Formula Used:
Sum to 100 terms is $S=\sum\limits_{r=1}^{100}{\left[ 1/\left[ {{(2r+1)}^{2}}-1 \right] \right]}$

Complete step by step Solution:
Given that
$1 /\left(3^{2}-1\right)+1 /\left(5^{2}-1\right)+1 /\left(7^{2}-1\right)+\ldots \ldots . .+1 /\left(201^{2}-1\right)$
Each number or component of a series is multiplied by the same number to create a geometric progression, or G.P. The constant ratio is this value. The ratio between any two successive numbers in a G.P. is the same number, which we refer to as the constant ratio. The letter "r" is typically used to represent it. Therefore, the ratio of any two consecutive numbers in a series, let's say a1, a2, a3,..., an, will be the same.
Sum upto 100 terms is defined as
$S=\sum\limits_{r=1}^{100}{\left[ 1/\left[ {{(2r+1)}^{2}}-1 \right] \right]}$
$S=\sum_{r=1}^{100}\left[1 /\left[(2 r+1)^{2}-1\right]\right]=\sum_{r=1}^{100} 1 /[(2 r+2)(2 r)]$
simillarly
In the expression we are substituting the values for $S=\sum_{r=1}^{100}\left[1 /\left[(2 r+1)^{2}-1\right]\right]=\sum_{r=1}^{100} 1 /[(2 r+2)(2 r)]$
In a GP, the common ratio is the common multiple between each succeeding phrase and the one before it. Each term in the geometric series multiplies a constant value to produce the following term. If the first term is $a$ and the second word is $ar$
We know that
according to cross multiplication
$(1/100)-(1/101)=(1/{{201}^{2}}-1)$
Then
$S=(1 / 4)[(1-(1 / 2))+((1 / 2)-(1 / 3))+((1 / 3)-(1 / 4) \ldots \ldots((1 / 100)-(1 / 101))$
Simplifying terns
$S=(1 / 4)(100 / 101)$
$=25 / 101$

Hence, the correct option is d.

Note: When one term is varied by another by a common ratio, the series is referred to as a geometric sequence. When we multiply the previous term by a constant (which is non-zero), we get the following term in the sequence.