Exponential Function MCQ Quiz - Objective Question with Answer for Exponential Function - Download Free PDF

Calculation:

${\displaystyle \Rightarrow \underset{\mathrm{x}\to \frac{{\pi }^{+}}{2}}{lim}{\mathrm{e}}^{\frac{\cot 6\mathrm{x}}{\cot 4\mathrm{x}}}=\underset{\mathrm{x}\to \frac{{\pi }^{+}}{2}}{lim}{\mathrm{e}}^{\frac{\sin 4\mathrm{x}\cdot \cos 6\mathrm{x}}{\sin 6\mathrm{x}\cdot \cos 4\mathrm{x}}}={\mathrm{e}}^{2/3}}$

${\displaystyle \Rightarrow \underset{\mathrm{x}\to \frac{{\pi }^{-}}{2}}{lim}(1+|\cos \mathrm{x}|{)}^{\frac{\lambda }{\cos \mathrm{x}}|}={\mathrm{e}}^{\lambda }}$

⇒ f(π/2) = µ 

For continuous function ⇒ e^2/3 = e^λ = µ 

$\lambda =\frac{2}{3},\mu ={\mathrm{e}}^{2/3}$

Now, 6λ + 6log_eµ + µ⁶ – e^6λ = 10

Hence, the correct answer is Option 4. 

Concept Used:-

We know that the Taylor expansion of e^x,

$e^x=1+\frac{x}{1}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \mathrm{\infty }$

On putting x = 1 we get,

$$\begin{gathered} \Rightarrow e=1+1+\frac{1}{2!}+\frac{1}{3!}+.... \\ \Rightarrow e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.... \\ \Rightarrow e=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n)!}.....(1) \end{gathered}$$

Also,

 $$\begin{gathered} \Rightarrow e=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n-1)!}.....(2) \\ \Rightarrow e=\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n-2)!}.....(3) \end{gathered}$$

Explanation:-

Given series is,

$1+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\dots \dots .$

Since, 0! is equal to 1. So, the above sum can be written as,

$\frac{1}{0!}+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\dots \dots .$

Let the sum of the above series is S. Then,

$S=\frac{1^2}{0!}+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\dots \dots .$

The denominator number written with factorial is one less than the numerator number written with square. Thus, in general, this series can be written as,

⇒ $S=\sum _{n=1}^n\frac{(n+1{)}^2}{(n)!}$

Expand the numerator (n+1)2 to solve it further,

$$\begin{gathered} \Rightarrow S=\sum _{n=0}^n\frac{(n+1{)}^2}{(n)!} \\ \Rightarrow S=\sum _{n=0}^n\frac{n^2+2n+1}{(n)!} \\ \Rightarrow S=\sum _{n=0}^n\frac{n^2}{(n)!}+\sum _{n=0}^n\frac{2n}{(n)!}+\sum _{n=0}^n\frac{1}{(n)!} \\ \Rightarrow S=\sum _{n=1}^n\frac{n(n-1)+n}{n.(n-1)!}+\sum _{n=1}^n\frac{2n}{n(.n-1)!}+\sum _{n=0}^n\frac{1}{(n)!} \\ \Rightarrow S=\sum _{n=2}^n\frac{1}{(n-2)!}+\sum _{n=1}^n\frac{1}{(n-1)!}+\sum _{n=1}^n\frac{2}{(n-1)!}+\sum _{n=0}^n\frac{1}{(n)!} \\ \Rightarrow S=\sum _{n=2}^n\frac{1}{(n-2)!}+\sum _{n=1}^n\frac{1}{(n-1)!}+2\sum _{n=1}^n\frac{1}{(n-1)!}+\sum _{n=0}^n\frac{1}{(n)!} \end{gathered}$$

Now using equation (1), (2) and (3) we get,

⇒ S = e + e + 2e + e

⇒ S = 5e

So, the sum of the given series is 5e.

Hence, the correct option is 5.

Calculation

Given

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{\mathrm{e}-(1+2\mathrm{x}{)}^{\frac{1}{2\mathrm{x}}}}{\mathrm{x}}}$

⇒ ${\mathrm{Lim}}_{x\to 0}\frac{e-e^{\frac{1}{2x}\ln (1+2x)}}{x}$

⇒  ${\mathrm{Lim}}_{x\to 0}(-e)\frac{(e^{\frac{\ln (1+2x)}{2x}-1}-1)}{x}$

⇒  ${\mathrm{Lim}}_{x\to 0}(-e)\frac{\ln (1+2x)-2x}{2x^2}$

⇒  $(-\mathrm{e})\times (-1)\frac{4}{2\times 2}=\mathrm{e}$ 

Hence option (1) is correct

Calculation

Given $\underset{x\to 0}{lim}\frac{(1+x{)}^{\frac{1}{x}}-e+\frac{1}{2}ex}{x^2}$

Let y = $(1+x{)}^{\frac{1}{x}}$

⇒ $\log y=\frac{1}{x}\log (1+x)=\frac{1}{x}[x=\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots .]$

⇒ logy = $1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\dots$

⇒ y = $e^{(1-\frac{x}{2}+\frac{x^2}{3}+..)}=e{e}^{(-\frac{x}{2}+\frac{x^2}{3}-)}$

⇒ y = $e[1+(-\frac{x}{2}+\frac{x^2}{3}-\dots )+\frac{1}{2!}{(-\frac{x}{2}+\frac{x^2}{3}-\dots )}^2+\dots ]$

⇒ $y-e+\frac{1}{2}ex=e{x}^2[\frac{1}{3}+0(x)+\frac{1}{2}{(-\frac{1}{2}+0(x))}^2+\dots ]$

[∵ 0(x) in terms containing x]

$\underset{x\to 0}{lim}\frac{y-e+\frac{1}{2}ex}{x^2}=e[\frac{1}{3}+\frac{1}{8}]=\frac{11}{24}e$

Hence option 1 is correct

Calculation

Given

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{\mathrm{e}-(1+2\mathrm{x}{)}^{\frac{1}{2\mathrm{x}}}}{\mathrm{x}}}$

⇒ ${\mathrm{Lim}}_{x\to 0}\frac{e-e^{\frac{1}{2x}\ln (1+2x)}}{x}$

⇒  ${\mathrm{Lim}}_{x\to 0}(-e)\frac{(e^{\frac{\ln (1+2x)}{2x}-1}-1)}{x}$

⇒  ${\mathrm{Lim}}_{x\to 0}(-e)\frac{\ln (1+2x)-2x}{2x^2}$

⇒  $(-\mathrm{e})\times (-1)\frac{4}{2\times 2}=\mathrm{e}$ 

Hence option (1) is correct

Concept:

$\underset{\mathrm{x}\to \mathrm{a}}{lim}\left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{g}\left(\mathrm{x}\right)\right]=\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}\left(\mathrm{x}\right)+\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{g}\left(\mathrm{x}\right)$

$\underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{({\mathrm{a}}^{\mathrm{x}}-1)}{\mathrm{x}}}=\log \mathrm{a}$

log mⁿ = n log m

Calculation:

$$\begin{gathered} {\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{3^{\mathrm{x}}+3^{-\mathrm{x}}-2}{\mathrm{x}}}} \\ ={\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{3^{\mathrm{x}}-1+3^{-\mathrm{x}}-1}{\mathrm{x}}}} \\ ={\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{3^{\mathrm{x}}-1}{\mathrm{x}}}+{\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{3^{-\mathrm{x}}-1}{\mathrm{x}}}}} \\ ={\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{3^{\mathrm{x}}-1}{\mathrm{x}}}+{\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{(3^{-1}{)}^{\mathrm{x}}-1}{\mathrm{x}}}}} \\ =\log 3+\log (3^{-1}) \\ =\log 3-\log 3 \\ =0 \end{gathered}$$

Concept:

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}({\mathrm{a}}^{\mathrm{x}})={\mathrm{a}}^{\mathrm{x}}\ln \mathrm{a}$

When there is 0/0 or ∞/∞ form exist we use L-Hospital Rule according to which we differentiate numerator and denominator separately.

Calculation:

$\underset{\mathrm{x}\to 0}{lim}\frac{5^{\mathrm{x}}-1}{\mathrm{x}}$

0/0 form, So using L-Hospital rule, we get 

$\begin{array}{l}\underset{x\to 0}{lim}\frac{5^x\ln 5-0}{1}\\ ={\log }_{e}5\end{array}$

Hence, option (1) is correct.

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{0}{0}$

II. $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$

Then we can apply L-Hospital Rule ⇔ $\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{\mathbf{f}\left(\mathbf{x}\right)}{\mathbf{g}\left(\mathbf{x}\right)}=\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{{\mathbf{f}}^{\prime }\left(\mathbf{x}\right)}{{\mathbf{g}}^{\prime }\left(\mathbf{x}\right)}$ 

Note: We have to differentiate both the numerator and denominator with respect to x unless and until $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\mathrm{l}\ne \frac{0}{0}$ where l is a finite value.

Calculation:

$\underset{x\to 0}{lim}\frac{e^x-\left(1+x\right)}{x^2}$                 [Form 0/0]

Apply L-Hospital Rule,

$=\underset{x\to 0}{lim}\frac{e^x-\left(0+1\right)}{2x}=\underset{x\to 0}{lim}\frac{e^x-1}{2x}$              [Form 0/0]

Again apply L-Hospital Rule,

$=\underset{x\to 0}{lim}\frac{e^x-0}{2}=\frac{e^0}{2}=\frac{1}{2}$

Given:

f(x) = ${\displaystyle \underset{\mathrm{n}\to \infty }{lim}\frac{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}{{\mathrm{a}}^{\mathrm{n}}-{\mathrm{b}}^{\mathrm{n}}}}$ and

a > b > 1

Calculation:

We have,

a > b > 1

⇒ $\frac{a}{b}>1$ or $\frac{b}{a}<1$

Given that,

f(x) = ${\displaystyle \underset{\mathrm{n}\to \infty }{lim}\frac{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}{{\mathrm{a}}^{\mathrm{n}}-{\mathrm{b}}^{\mathrm{n}}}}$    

⇒ f(x) = ${\displaystyle \underset{\mathrm{n}\to \infty }{lim}\frac{{\mathrm{a}}^{\mathrm{n}}[1+(\frac{\mathrm{b}}{\mathrm{a}}{)}^{\mathrm{n}}]}{{\mathrm{a}}^{\mathrm{n}}[1-(\frac{\mathrm{b}}{\mathrm{a}}{)}^{\mathrm{n}}]}}$ 

⇒ f(x) = ${\displaystyle \underset{\mathrm{n}\to \infty }{lim}\frac{[1+(\frac{\mathrm{b}}{\mathrm{a}}{)}^{\mathrm{n}}]}{[1-(\frac{\mathrm{b}}{\mathrm{a}}{)}^{\mathrm{n}}]}}$ 

Taking limit n→∞

⇒ f(x) = ${\displaystyle \underset{\mathrm{n}\to \infty }{lim}\frac{[1+(\frac{{\mathrm{b}}^{\mathrm{\infty }}}{{\mathrm{a}}^{\mathrm{\infty }}})]}{[1-(\frac{{\mathrm{b}}^{\mathrm{\infty }}}{{\mathrm{a}}^{\mathrm{\infty }}})]}}$  

⇒ f(x) = $\frac{1+0}{1-0}$       (∵ $\frac{b}{a}<1$ )

∴  f(x) = 1

Given:

The given limiting function is as follows,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}-{\mathrm{x}}^2}{{\mathrm{x}}^2}}$ 

Concept:

Use the L'hospital rule for the limit $\underset{x\to a}{lim}\frac{f(x)}{g(x)}$ of the form $\frac{0}{0}$

$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Where,

f'(x) and g'(x) are the derivative of f(x) and g(x).

Solution:

The given limit is as follows,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}-{\mathrm{x}}^2}{{\mathrm{x}}^2}}$ 

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}-{\mathrm{x}}^2}{{\mathrm{x}}^2}=\frac{0}{0}}$

Derivating f(x) and g(x) we will get,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-1-2\mathrm{x}}{2\mathrm{x}}=\frac{0}{0}}$

Still, it is of $\frac{0}{0}$ form.

Derivating again we will get,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-2}{2}=\frac{1-2}{2}=\frac{-1}{2}}$

Hence, option 3 is correct.

Concept:

Step: 1

Check f(a) is defined. If it is not defined then no need to go further.  The function is not continuous at a. If f(a) is defined then

Step: 2

Check Left-hand limit (LHL) and Right-hand limit (RHL) 

$LHL=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}(\mathrm{x})$

$RHL=\underset{x\to a^{+}}{lim}f(x)$

If LHL = RHL then limit exist.

Formula used:

1) 1 - cos 2θ = 2sin²θ 

2) ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\left(\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}{\mathrm{x}}\right)}$ = 1

Calculation: 

We have ${\displaystyle \underset{\mathrm{x}\to 1}{lim}\left(\frac{\surd 1-\mathrm{c}\mathrm{o}\mathrm{s}2(\mathrm{x}-1)}{(\mathrm{x}-1)}\right)}$

By using the above formula 

${\displaystyle \underset{\mathrm{x}\to 1}{lim}\left(\frac{\surd 2\mathrm{s}\mathrm{i}{\mathrm{n}}^2(\mathrm{x}-1)}{(\mathrm{x}-1)}\right)}$

$\surd 2{\displaystyle \underset{\mathrm{x}\to 1}{lim}\left(\frac{|\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{x}-1)|}{(\mathrm{x}-1)}\right)}$

Since there is a mod in the expression, we have to check, whether the limit exists or not. 

LHL:

$\surd 2{\displaystyle \underset{\mathrm{x}\to 1^{-}}{lim}\left(\frac{|\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{x}-1)|}{(\mathrm{x}-1)}\right)}$

$-\surd 2{\displaystyle \underset{\mathrm{x}\to 1^{-}}{lim}\left(\frac{\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{x}-1)}{(\mathrm{x}-1)}\right)}$

By using the formula (2)

$-\surd 2{\displaystyle \underset{\mathrm{x}-1\to 0^{}}{lim}\left(\frac{\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{x}-1)}{(\mathrm{x}-1)}\right)}$

LHL = -√2

RHL: 

$\surd 2{\displaystyle \underset{\mathrm{x}\to 1^{+}}{lim}\left(\frac{|\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{x}-1)|}{(\mathrm{x}-1)}\right)}$

RHL = √2

Since, LHL ≠ RHL

The limit of the expression does not exist.

CONCEPT:

• $\underset{x\to 0}{lim}\left[\frac{a^x-1}{x}\right]=\log a,a>0$
• $\underset{x\to 0}{lim}\left[\frac{e^x-1}{x}\right]=1$
• $\underset{x\to a}{lim}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to a}{lim}f\left(x\right)}{\underset{x\to a}{lim}g\left(x\right)},provided\underset{x\to a}{lim}g\left(x\right)\ne 0$

CALCULATION:

Here, we have to find the limit of ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{a^x-b^x}{e^x-1}}}$

The expression $\frac{a^x-b^x}{e^x-1}$ can be re-written as:

$\Rightarrow \frac{a^x-b^x}{e^x-1}=\left[\frac{\frac{a^x-1}{x}-\frac{b^x-1}{x}}{\frac{e^x-1}{x}}\right]$

Now by applying limits on both the sides of the above equation we get

$\Rightarrow \underset{x\to 0}{lim}\frac{a^x-b^x}{e^x-1}=\underset{x\to 0}{lim}\left[\frac{\frac{a^x-1}{x}-\frac{b^x-1}{x}}{\frac{e^x-1}{x}}\right]$

As we know that, $\underset{x\to a}{lim}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\underset{x\to a}{lim}f\left(x\right)}{\underset{x\to a}{lim}g\left(x\right)},provided\underset{x\to a}{lim}g\left(x\right)\ne 0$

$\Rightarrow \underset{x\to 0}{lim}\left[\frac{\frac{a^x-1}{x}-\frac{b^x-1}{x}}{\frac{e^x-1}{x}}\right]=\frac{\left[\underset{x\to 0}{lim}\frac{a^x-1}{x}\right]-\left[\underset{x\to 0}{lim}\frac{b^x-1}{x}\right]}{\left[\underset{x\to }{lim}\frac{e^x-1}{x}\right]}$

As we know that, $\underset{x\to 0}{lim}\left[\frac{a^x-1}{x}\right]=\log a,a>0$ and $\underset{x\to 0}{lim}\left[\frac{e^x-1}{x}\right]=1$

$\Rightarrow \underset{x\to 0}{lim}\frac{a^x-b^x}{e^x-1}=\log \frac{a}{b}$

Hence, option A is true.

Concept Used:-

We know that the Taylor expansion of e^x,

$e^x=1+\frac{x}{1}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \mathrm{\infty }$

On putting x = 1 we get,

$$\begin{gathered} \Rightarrow e=1+1+\frac{1}{2!}+\frac{1}{3!}+.... \\ \Rightarrow e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.... \\ \Rightarrow e=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n)!}.....(1) \end{gathered}$$

Also,

 $$\begin{gathered} \Rightarrow e=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n-1)!}.....(2) \\ \Rightarrow e=\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n-2)!}.....(3) \end{gathered}$$

Explanation:-

Given series is,

$1+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\dots \dots .$

Since, 0! is equal to 1. So, the above sum can be written as,

$\frac{1}{0!}+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\dots \dots .$

Let the sum of the above series is S. Then,

$S=\frac{1^2}{0!}+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\dots \dots .$

The denominator number written with factorial is one less than the numerator number written with square. Thus, in general, this series can be written as,

⇒ $S=\sum _{n=1}^n\frac{(n+1{)}^2}{(n)!}$

Expand the numerator (n+1)2 to solve it further,

$$\begin{gathered} \Rightarrow S=\sum _{n=0}^n\frac{(n+1{)}^2}{(n)!} \\ \Rightarrow S=\sum _{n=0}^n\frac{n^2+2n+1}{(n)!} \\ \Rightarrow S=\sum _{n=0}^n\frac{n^2}{(n)!}+\sum _{n=0}^n\frac{2n}{(n)!}+\sum _{n=0}^n\frac{1}{(n)!} \\ \Rightarrow S=\sum _{n=1}^n\frac{n(n-1)+n}{n.(n-1)!}+\sum _{n=1}^n\frac{2n}{n(.n-1)!}+\sum _{n=0}^n\frac{1}{(n)!} \\ \Rightarrow S=\sum _{n=2}^n\frac{1}{(n-2)!}+\sum _{n=1}^n\frac{1}{(n-1)!}+\sum _{n=1}^n\frac{2}{(n-1)!}+\sum _{n=0}^n\frac{1}{(n)!} \\ \Rightarrow S=\sum _{n=2}^n\frac{1}{(n-2)!}+\sum _{n=1}^n\frac{1}{(n-1)!}+2\sum _{n=1}^n\frac{1}{(n-1)!}+\sum _{n=0}^n\frac{1}{(n)!} \end{gathered}$$

Now using equation (1), (2) and (3) we get,

⇒ S = e + e + 2e + e

⇒ S = 5e

So, the sum of the given series is 5e.

Hence, the correct option is 1.

Concept Used:-

It is known that the derivative of eˣ is eˣ.

Now when it is written that,

eˣ = a₀ + a₁ x + a₂x² + ... + a_nxⁿ

At, x = 0; a₀ = 1

Differentiating the above equation we get,

eˣ = a₁ + 2a₂x + ...+ na_nx^n-1

Now put x = 0; a₁ = 1

Differentiate it again,

eˣ = 2a₂ +... n (n - 1)x^n-2

Now at, x = 0, 2a₂ = 1

When we, again and again, differentiate it, and put x = 0 to solve for the value of a_n, we get the series:

1 + x + x²/2! + x³/3! + ...+ xⁿ/n! to infinity.

On putting x = 1 we have,

e = 1 + 1/1! + 1/2! + 1/3! + ...+ 1/n!              .......(1)

Explanation:-

Given that the nth term of a series is $\frac{1}{(\mathrm{n}+1)!}$.

$T_n=\frac{1}{(\mathrm{n}+1)!}$

Here, the series goes from n = 1 to ∞. So, the series will be,

$S=\frac{1}{2!}+\frac{1}{3!}+\dots -\mathrm{\infty }$

Here, add and subtract 1+1/1! in the right-hand side,

$$\begin{gathered} \Rightarrow S=(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+--\mathrm{\infty })-(1+\frac{1}{1!}) \\ \Rightarrow S=(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+--\mathrm{\infty })-2 \end{gathered}$$

From equation (1) we get, 

⇒ S = e - 2

So, the sum of the series will be (e - 2)

Formula used:

X(a + b) = X^c then, (a + b) = c

Calculation:

Let 2x = 3y = 6^+z = k 

2x = k then, 2 = k^1/x

3^y = k then, 3 = k1/y

6^+z = k then, 6 = k1/z 

Now, We know that

2 × 3 = 6

⇒ k^1/x × k^1/y = k^1/z

⇒ k^{(1/x + 1/y)} = k^1/z

⇒ 1/x + 1/y = 1/z

⇒ -(1/x + 1/y) + 1/z = 0 

Note:- The official question is wrong, the updated question and the solution is provided.