Question

FInd the value of $\dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} + \dfrac{{{C_6}}}{7}........... =$

Answer 1

Use binomial expansions of ${\left( {1 + x} \right)^n}$ and ${\left( {1 - x} \right)^n}$ and then add them so that the terms that you don’t need will be cancel out. After doing this integrate both sides of the equation by using the formula that integral of ${x^n}$ is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$. Therefore on solving further we will be able to find the value of the given expression.

Complete step-by-step solution:
In the question, they are asking us for the value of $\dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} + \dfrac{{{C_6}}}{7}...........$.
We know that the binomial expansion of ${\left( {1 + x} \right)^n} = {}^n{C_0}{x^0} + {}^n{C_1}{x^1} + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} +............ + {}^n{C_n}{x^n}$. On further solving we get
${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} +............ + {}^n{C_n}{x^n}$ ----------- (i)
Similarly, the binomial expansion of ${\left( {1 - x} \right)^n}$ will be
${\left( {1 - x} \right)^n} = {}^n{C_0} - {}^n{C_1}x + {}^n{C_2}{x^2} - {}^n{C_3}{x^3} +............ + {\left( { - 1} \right)^n}{}^n{C_n}{x^n}$ ------------- (ii)
We know that we have to find the value of $\dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} + \dfrac{{{C_6}}}{7}...........$. In this expression ${C_1}$ , ${C_3}$ , ${C_5}$, ………. are absent. That means we only need the values of ${C_0}$ , ${C_2}$ , ${C_4}$ , ${C_6}$ ,………... Therefore we will now add the equations (i) and (ii). By adding these two equations we get
${\left( {1 + x} \right)^n} + {\left( {1 - x} \right)^n} = \left( {{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} +............ + {}^n{C_n}{x^n}} \right) + \left( {{}^n{C_0} - {}^n{C_1}x + {}^n{C_2}{x^2} - {}^n{C_3}{x^3} +............ + {{\left( { - 1} \right)}^n}{}^n{C_n}{x^n}} \right)$
Now cancel out the similar terms with opposite signs. On doing this we get
${\left( {1 + x} \right)^n} + {\left( {1 - x} \right)^n} = 2\left( {{}^n{C_0} + {}^n{C_2}{x^2} + {}^n{C_4}{x^4} +...........} \right)$
The given expression has a denominator in every term. Therefore we will now integrate the above equation. On integrating both sides we get
$\int {\left( {{{\left( {1 + x} \right)}^n} + {{\left( {1 - x} \right)}^n}} \right)} dx = 2\int {{}^n{C_0} + {}^n{C_2}{x^2} + {}^n{C_4}{x^4} +...........}$
First we will integrate the right hand side to avoid the confusion. We know that the integral of ${x^n}$ is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$. Therefore the above expression will be
$\int {\left( {{{\left( {1 + x} \right)}^n} + {{\left( {1 - x} \right)}^n}} \right)} dx = 2\left( {{C_0}.\dfrac{{{x^{0 + 1}}}}{{0 + 1}} + {C_2}.\dfrac{{{x^{2 + 1}}}}{{2 + 1}} + {C_4}\dfrac{{{x^{4 + 1}}}}{{4 + 1}} +...........} \right)$
On further solving we get
$\int {\left( {{{\left( {1 + x} \right)}^n} + {{\left( {1 - x} \right)}^n}} \right)} dx = 2\left( {\dfrac{{{C_0}}}{1}.x + \dfrac{{{C_2}}}{3}.{x^3} + \dfrac{{{C_4}}}{5}.{x^5} +...........} \right)$
Now the same formula we will apply in the left hand side that we used for the right hand side to find the integral. Therefore on integrating left hand side we get
$\dfrac{{{{\left( {1 + x} \right)}^{n + 1}}}}{{n + 1}} + \left( { - \dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}} \right) = 2\left( {\dfrac{{{C_0}}}{1}.x + \dfrac{{{C_2}}}{3}.{x^3} + \dfrac{{{C_4}}}{5}.{x^5} +...........} \right)$
On putting x is equal to one on both sides we get
$\dfrac{{{{\left( {1 + 1} \right)}^{n + 1}}}}{{n + 1}} - \dfrac{{{{\left( {1 - 1} \right)}^{n + 1}}}}{{n + 1}} = 2\left( {\dfrac{{{C_0}}}{1}.1 + \dfrac{{{C_2}}}{3}.{{\left( 1 \right)}^3} + \dfrac{{{C_4}}}{5}.{{\left( 1 \right)}^5} +...........} \right)$
By doing further calculations we get
$\dfrac{{{2^{n + 1}}}}{{n + 1}} - \dfrac{{{{\left( 0 \right)}^{n + 1}}}}{{n + 1}} = 2\left( {\dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} +...........} \right)$
That is
$\dfrac{{{2^{n + 1}}}}{{n + 1}} - 0 = 2\left( {\dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} +...........} \right)$
On shifting two from the right hand side to the left hand side we get
$\dfrac{{{2^{n + 1}}}}{{2\left( {n + 1} \right)}} = \dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} +...........$
We can also write it as
$\dfrac{{{2^{n + 1}}{{.2}^{ - 1}}}}{{\left( {n + 1} \right)}} = \dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} +...........$
On adding powers of two we get
$\dfrac{{{2^n}}}{{n + 1}} = \dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} +...........$
Hence the value of $\dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} +........... = \dfrac{{{2^n}}}{{n + 1}}$.

Note: Remember that ${}^n{C_0}$ , ${}^n{C_1}$ , ${}^n{C_2}$ , ${}^n{C_3}$ ,……….. ${}^n{C_n}$ are called binomial coefficients and also represented by ${C_0}$ , ${C_1}$ , ${C_2}$ , ${C_3}$ ,……….. ${C_n}$ . Keep in mind that the integral of ${\left( {1 - x} \right)^n}$ is $- \dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}$ . Here the reason for the negative sign of the integral is because the x is negative. Check the steps carefully for better understanding.