Question

If we have an expression as $\dfrac{{{e^x}}}{{1 - x}} = {B_o} + {B_1}x + {B_2}{x^2} + ...... + {B_{n - 1}}{x^{n - 1}} + {B_n}{x^n}$ then ${B_n} - {B_{n - 1}}$ equals
$\left( a \right)\dfrac{1}{{n!}}$
$\left( b \right)\dfrac{1}{{\left( {n - 1} \right)!}}$
$\left( c \right)\dfrac{1}{{n!}} - \dfrac{1}{{\left( {n - 1} \right)!}}$
$\left( d \right)1$

Answer 1

In this particular question use the concept of expansion and then multiplication, the expansion of ${e^x}$ is $\left( {1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + ..... + \dfrac{{{x^{n - 1}}}}{{\left( {n - 1} \right)!}} + \dfrac{{{x^n}}}{{n!}}} \right)$ and the expansion of ${\left( {1 - x} \right)^{ - 1}}$ is given as, $\left( {1 + x + {x^2} + {x^3} + ..... + {x^{n - 1}} + {x^n}} \right)$ so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given equation:
$\dfrac{{{e^x}}}{{1 - x}} = {B_o} + {B_1}x + {B_2}{x^2} + ...... + {B_{n - 1}}{x^{n - 1}} + {B_n}{x^n}$
Now we have to find out the value of ${B_n} - {B_{n - 1}}$
Now consider the LHS of the given equation we have,
$\Rightarrow \dfrac{{{e^x}}}{{1 - x}}$
Now the above equation is written as
$\Rightarrow {e^x}{\left( {1 - x} \right)^{ - 1}}$
Now as we know that the expansion of ${e^x}$ is $\left( {1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + ..... + \dfrac{{{x^{n - 1}}}}{{\left( {n - 1} \right)!}} + \dfrac{{{x^n}}}{{n!}}} \right)$ and the expansion of ${\left( {1 - x} \right)^{ - 1}}$ is given as, $\left( {1 + x + {x^2} + {x^3} + ..... + {x^{n - 1}} + {x^n}} \right)$ so we have,
$\Rightarrow {e^x}{\left( {1 - x} \right)^{ - 1}} = \left( {1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + ..... + \dfrac{{{x^{n - 1}}}}{{\left( {n - 1} \right)!}} + \dfrac{{{x^n}}}{{n!}}} \right)\left( {1 + x + {x^2} + {x^3} + ..... + {x^{n - 1}} + {x^n}} \right)$

$$\Rightarrow {e^x}{\left( {1 - x} \right)^{ - 1}} = 1 + \left( {1 + \dfrac{1}{{1!}}} \right)x + \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}}} \right){x^2} + \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}}} \right){x^3} + .... + \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + ... + \dfrac{1}{{\left( {n - 1} \right)!}}} \right){x^{n - 1}} \\\ \+ \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + ... + \dfrac{1}{{\left( {n - 1} \right)!}} + \dfrac{1}{{n!}}} \right){x^n} \\\$$

Now compare the coefficients of the RHS of the above equation with the RHS of the given equation we have,
$\dfrac{{{e^x}}}{{1 - x}} = {B_o} + {B_1}x + {B_2}{x^2} + ...... + {B_{n - 1}}{x^{n - 1}} + {B_n}{x^n}$
$\Rightarrow {B_o} = 1,{B_1} = \left( {1 + \dfrac{1}{{1!}}} \right),......,{B_{n - 1}} = \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + ... + \dfrac{1}{{\left( {n - 1} \right)!}}} \right),{B_n} = \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + ... + \dfrac{1}{{\left( {n - 1} \right)!}} + \dfrac{1}{{n!}}} \right)$
Now we have to find out the value of ${B_n} - {B_{n - 1}}$.
$\Rightarrow {B_n} - {B_{n - 1}} = \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + ... + \dfrac{1}{{\left( {n - 1} \right)!}} + \dfrac{1}{{n!}}} \right) - \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + ... + \dfrac{1}{{\left( {n - 1} \right)!}}} \right)$
So as we see that all the terms are cancel out except one term so we have,
$\Rightarrow {B_n} - {B_{n - 1}} = \dfrac{1}{{n!}}$
So this is the required answer.
Hence option (a) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the expansion of standard terms such as ${e^x}{\text{ and }}{\left( {1 - x} \right)^{ - 1}}$ which is all written above then multiply them as above and write the coefficients of ${B_o},{B_1},{B_2}......{B_{n - 1}},{B_n}$ as above then find out the value of $\left( {{B_n} - {B_{n - 1}}} \right)$ as above, we will get the required answer.