Question

The coefficient of ${x^n}$, where n is any positive integer, in the expansion of ${\left( {1 + 2x + 3{x^2} + ..... + \infty } \right)^{\dfrac{1}{2}}}$ is
$\left( a \right)$ 1
$\left( b \right)\dfrac{{n + 1}}{2}$
$\left( c \right)$ 2n + 1
$\left( d \right)$ n + 1

Answer 1

Hint: In this particular type of question use the concept that the binomial expansion of ${\left( {1 + x} \right)^n}$ is given as, ${}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + .............. + {}^n{C_n}{x^n}$, so in place of n substitute, -2 and in place of x substitute –x, so use these concepts to get the solution of the question.

Complete step-by-step answer:
Given equation,
${\left( {1 + 2x + 3{x^2} + ..... + \infty } \right)^{\dfrac{1}{2}}}$................ (1)
According to binomial theorem the expansion of ${\left( {1 + x} \right)^n}$ is
$ \Rightarrow {\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + .............. + {}^n{C_n}{x^n}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property in above equation we have,
$ \Rightarrow {}^n{C_0} = \dfrac{{n!}}{{0!\left( {n - 0} \right)!}} = 1$, ${}^n{C_1} = \dfrac{{n!}}{{1!\left( {n - 1} \right)!}} = n$, ${}^n{C_2} = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = \dfrac{{n\left( {n - 1} \right)}}{{2!}}$ , ${}^n{C_3} = \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}$ and so on......., so the above equation converts into,
$ \Rightarrow {\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ..............$ ............... (2)
Now substitute in place of n, -2 and in place of x, -x in the above equation we have,
$ \Rightarrow {\left( {1 - x} \right)^{ - 2}} = 1 + \left( { - 2} \right)\left( { - x} \right) + \dfrac{{\left( { - 2} \right)\left( { - 2 - 1} \right)}}{{2!}}{\left( { - x} \right)^2} + \dfrac{{\left( { - 2} \right)\left( { - 2 - 1} \right)\left( { - 2 - 2} \right)}}{{3!}}{\left( { - x} \right)^3} + .............. + \infty $
Now simplify we have,
$ \Rightarrow {\left( {1 - x} \right)^{ - 2}} = 1 + 2x + 3{x^2} + 4{x^3} + .............. + \infty $
Now substitute this value in equation (1) we have,
\[ \Rightarrow {\left( {1 + 2x + 3{x^2} + ..... + \infty } \right)^{\dfrac{1}{2}}} = {\left( {{{\left( {1 - x} \right)}^{ - 2}}} \right)^{\dfrac{1}{2}}}\]
\[ \Rightarrow {\left( {1 + 2x + 3{x^2} + ..... + \infty } \right)^{\dfrac{1}{2}}} = {\left( {1 - x} \right)^{ - 1}}\]....................... (3)
Now substitute n = -1 and x = -x in equation (2) we have,
$ \Rightarrow {\left( {1 - x} \right)^{ - 1}} = 1 + \left( { - 1} \right)\left( { - x} \right) + \dfrac{{\left( { - 1} \right)\left( { - 1 - 1} \right)}}{{2!}}{\left( { - x} \right)^2} + \dfrac{{\left( { - 1} \right)\left( { - 1 - 1} \right)\left( { - 1 - 2} \right)}}{{3!}}{\left( { - x} \right)^3} + .............. + \infty $
Now simplify we have,
$ \Rightarrow {\left( {1 - x} \right)^{ - 1}} = 1 + x + {x^2} + {x^3} + .............. + {x^n} + ............\infty $
So from equation (3) we have,
\[ \Rightarrow {\left( {1 + 2x + 3{x^2} + ..... + \infty } \right)^{\dfrac{1}{2}}} = 1 + x + {x^2} + {x^3} + .............. + {x^n} + ............\infty \]
Now as we see from the above equation the coefficient of \[{x^n}\] in the expansion of ${\left( {1 + 2x + 3{x^2} + ..... + \infty } \right)^{\dfrac{1}{2}}}$ is 1.
So this is the required answer.
Hence option (A) is the correct answer.

Note – Whenever we face such types of question the key concept we have to remember is that always recall the binomial expansion and the formula of combination which is all stated above, so first use the expansion formula as above then apply the combination formula as above then use the expansion formula as above and check what is the coefficient of ${x^n}$ which is the required answer.