Question

Find the sum of the series $\dfrac{2}{{1!}} + \dfrac{{(2 + 4)}}{{2!}} + \dfrac{{(2 + 4 + 6)}}{{3!}} + ....\infty $.
A.$e$
B. 2e
C. 3e
D. None of these

Answer 1

Hint: Write 2 as 1.2, (2+4) that is 6 as 3.2, (2+4+6) that is 12 as 3.4, (2+4+6+8) that is 20 as 4.5 in the numerators of the given series. Then simplify the series. Now, write the series of ${e^x}$, then multiply both sides of the equation ${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....$ by ${x^2}$. Differentiate both sides of the result with respect to x. At last substitute 1 for x in the differentiated result to obtain the required answer.

Formula Used:
$n! = 1.2.3....n$
$n! = n(n - 1)!$
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....$
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$

Complete step by step solution:
The given series can be written as,
$\dfrac{2}{{1!}} + \dfrac{6}{{2!}} + \dfrac{{12}}{{3!}} + \dfrac{{20}}{{4!}} + ....\infty $
$\dfrac{{1.2}}{{1!}} + \dfrac{{2.3}}{{2!}} + \dfrac{{3.4}}{{3!}} + \dfrac{{4.5}}{{4!}} + ....\infty $
$\dfrac{{1.2}}{{1!}} + \dfrac{{2.3}}{{2.1!}} + \dfrac{{3.4}}{{3.2!}} + \dfrac{{4.5}}{{4.3!}} + ....\infty $
$2 + \dfrac{3}{{1!}} + \dfrac{4}{{2!}} + \dfrac{5}{{3!}} + ....\infty $
Now,
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....$
Multiply both sides of the equation ${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....$ by ${x^2}$ .
${x^2}{e^x} = {x^2} + \dfrac{{{x^3}}}{{1!}} + \dfrac{{{x^4}}}{{2!}} + \dfrac{{{x^5}}}{{3!}} + ....$
Differentiate both sides of the equation ${x^2}{e^x} = {x^2} + \dfrac{{{x^3}}}{{1!}} + \dfrac{{{x^4}}}{{2!}} + \dfrac{{{x^5}}}{{3!}} + ....$ with respect to x.
${x^2}{e^x} + {e^x}.2x = 2x + \dfrac{{3{x^2}}}{{1!}} + \dfrac{{4{x^3}}}{{2!}} + \dfrac{{5{x^4}}}{{3!}} + .....$
Substitute 1 for x in both sides of the equation ${x^2}{e^x} + {e^x}.2x = 2x + \dfrac{{3{x^2}}}{{1!}} + \dfrac{{4{x^3}}}{{2!}} + \dfrac{{5{x^4}}}{{3!}} + .....$
We get,
${1^2}{e^1} + {e^1}.2.1 = 2.1 + \dfrac{{{{3.1}^2}}}{{1!}} + \dfrac{{{{4.1}^3}}}{{2!}} + \dfrac{{{{5.1}^4}}}{{3!}} + .....$
$2 + \dfrac{3}{{1!}} + \dfrac{4}{{2!}} + \dfrac{5}{{3!}} + .... = 2e + e = 3e$

Option ‘C’ is correct

Note: To solve this type of question one needs to have the knowledge of the expansion of ${e^x}$, otherwise it will be difficult to solve this type of questions.