
Find $1.2 / 1 !+2.3 / 2 !+3.4 / 3 !+\ldots \infty$ is
A. $e$
B. $2 \mathrm{e}$
C. $3 \mathrm{e}$
D. none of these
Answer
163.2k+ views
Hint:Here we have to determine the number of factorials first. By applying the formula we can easily find the factorial. And after that, we have to put the values to find each term in the expansion. The main function of factorial is to count all possible distinct items up to "n".
Formula used:
The formula to determine a number's factorial is
$n !=n \times(n-1) \times(n-2) \times(n-3) \times \ldots \times 3 \times 2 \times 1$
Complete step by step Solution:
Given that
$1.2 / 1 !+2.3 / 2 !+3.4 / 3 !+\ldots \infty$
The factorial notation in terms of pi product nomenclature is as follows for an integer $n \geq 1$,:
$n !=\prod_{i=1}^{n} i$
According to the formulas above, the factorial of a number's recurrence relation is the sum of its factorial and its factorial less one. It comes from:
$n !=n .(n-1) !$
$\mathrm{T}_{\mathrm{n}}=\mathrm{n}(\mathrm{n}+1) / \mathrm{n} !$
$=(\mathrm{n}+1) /(\mathrm{n}-1) !$
Put the values as
$\text{n}=1,2,3,\ldots $
$\mathrm{T}_{1}=1.2 / 1 !=2 / 0 !$
$\mathrm{T}_{2}=1 / 0 !+2 / 1 !$
$\mathrm{T}_{3}=1 / 1 !+2 / 2 !$
$\mathrm{T}_{4}=1 / 2 !+2 / 3 !$
Re-arrange the values according to the equation
$\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}+\ldots . \infty$
$=(1 / 0 !+1 / 1 !+1 / 2 !+\ldots \infty)+2(1 / 0 !+1 / 1 !+1 / 2 !+\ldots \infty)$
Simplify the expression by substituting the values of each factorial
$=\mathrm{e}+2 \mathrm{e}(\mathrm{e}=1+1+1 / 2 !+1 / 3 !+1 / 4 !+1 / 5 !+\ldots .)$
$=3 \mathrm{e}$
Hence, the correct option is 3.
Additional information:
A number's factorial is the result of multiplying the integer by each natural number below it. Factorial can be symbolised by the letter "!". Thus, n factorial is denoted by n and is the result of the first n natural numbers!
Note:Note that the factorial is a crucial mathematical operation that is used to determine the number of possible arrangements or the ordered set of numbers. A factorial function, in essence, multiplies a number by each number below it until it equals 1. For instance, the factorial of 3 is equal to 6 and reflects the multiplication of the integers 3, 2, and 1.
Formula used:
The formula to determine a number's factorial is
$n !=n \times(n-1) \times(n-2) \times(n-3) \times \ldots \times 3 \times 2 \times 1$
Complete step by step Solution:
Given that
$1.2 / 1 !+2.3 / 2 !+3.4 / 3 !+\ldots \infty$
The factorial notation in terms of pi product nomenclature is as follows for an integer $n \geq 1$,:
$n !=\prod_{i=1}^{n} i$
According to the formulas above, the factorial of a number's recurrence relation is the sum of its factorial and its factorial less one. It comes from:
$n !=n .(n-1) !$
$\mathrm{T}_{\mathrm{n}}=\mathrm{n}(\mathrm{n}+1) / \mathrm{n} !$
$=(\mathrm{n}+1) /(\mathrm{n}-1) !$
Put the values as
$\text{n}=1,2,3,\ldots $
$\mathrm{T}_{1}=1.2 / 1 !=2 / 0 !$
$\mathrm{T}_{2}=1 / 0 !+2 / 1 !$
$\mathrm{T}_{3}=1 / 1 !+2 / 2 !$
$\mathrm{T}_{4}=1 / 2 !+2 / 3 !$
Re-arrange the values according to the equation
$\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}+\ldots . \infty$
$=(1 / 0 !+1 / 1 !+1 / 2 !+\ldots \infty)+2(1 / 0 !+1 / 1 !+1 / 2 !+\ldots \infty)$
Simplify the expression by substituting the values of each factorial
$=\mathrm{e}+2 \mathrm{e}(\mathrm{e}=1+1+1 / 2 !+1 / 3 !+1 / 4 !+1 / 5 !+\ldots .)$
$=3 \mathrm{e}$
Hence, the correct option is 3.
Additional information:
A number's factorial is the result of multiplying the integer by each natural number below it. Factorial can be symbolised by the letter "!". Thus, n factorial is denoted by n and is the result of the first n natural numbers!
Note:Note that the factorial is a crucial mathematical operation that is used to determine the number of possible arrangements or the ordered set of numbers. A factorial function, in essence, multiplies a number by each number below it until it equals 1. For instance, the factorial of 3 is equal to 6 and reflects the multiplication of the integers 3, 2, and 1.
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