Question

The value of \[1{}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + 7{}^n{C_3} + ...................................... + \left( {2n + 1} \right){}^n{C_n}\] is equal to
A. \[{2^n}\]
B. \[{2^n} + n{2^{n - 1}}\]
C. \[{2^n}\left( {n + 1} \right)\]
D. none of these

Answer 1

Hint: First of all, consider the expansion of \[{\left( {1 + {x^2}} \right)^n}\] by using binomial theorem. Then multiply both sides with \[x\] and differentiate w.r.t \[x\]. Then substitute \[x\] is equal to 1 to obtain the required answer.

Complete step-by-step answer:
As we know that
\[{\left( {1 + {x^2}} \right)^n} = 1{}^n{C_0} + {}^n{C_1}{x^2} + {}^n{C_2}{x^4} + {}^n{C_3}{x^6} + ................................................. + {}^n{C_n}{x^{2n}}\]
Multiplying both sides with \[x\], we get
\[
   \Rightarrow x{\left( {1 + {x^2}} \right)^n} = x\left( {1{}^n{C_0} + {}^n{C_1}{x^2} + {}^n{C_2}{x^4} + {}^n{C_3}{x^6} + ................................................. + {}^n{C_n}{x^{2n}}} \right) \\
   \Rightarrow x{\left( {1 + {x^2}} \right)^n} = x{}^n{C_0} + {}^n{C_1}{x^3} + {}^n{C_2}{x^5} + {}^n{C_3}{x^7} + ................................................. + {}^n{C_n}{x^{2n + 1}} \\
\]
Differentiating both sides with respect to \[x\], we get
\[\dfrac{d}{{dx}}\left[ {x{{\left( {1 + {x^2}} \right)}^n}} \right] = \dfrac{d}{{dx}}\left( {x{}^n{C_0} + {}^n{C_1}{x^3} + {}^n{C_2}{x^5} + {}^n{C_3}{x^7} + ................................................. + {}^n{C_n}{x^{2n + 1}}} \right)\]
We know that the product rule states that if \[f\left( x \right)\] and \[g\left( x \right)\] are both differentiable, then \[\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]\].
\[
   \Rightarrow x\dfrac{d}{{dx}}{\left( {1 + {x^2}} \right)^n} + {\left( {1 + {x^2}} \right)^n}\dfrac{d}{{dx}}\left( {x{}^n{C_0}} \right) = \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {{}^n{C_1}{x^3}} \right) + \dfrac{d}{{dx}}\left( {{}^n{C_2}{x^5}} \right) + \dfrac{d}{{dx}}\left( {{}^n{C_3}{x^7}} \right)........... + \dfrac{d}{{dx}}\left( {{}^n{C_n}{x^{2n + 1}}} \right) \\
   \Rightarrow x\left[ {n{{\left( {1 + {x^2}} \right)}^{n - 1}}\left( {2x} \right)} \right] + {\left( {1 + {x^2}} \right)^n}\left( 1 \right) = 1{}^n{C_0} + 3{}^n{C_1}{x^2} + 5{}^n{C_2}{x^4} + 7{}^n{C_3}{x^6} + .................. + \left( {2n + 1} \right){}^n{C_n}{x^{2n}} \\
\]
Now, put \[x = 1\]
\[
   \Rightarrow 1\left[ {n{{\left( {1 + {1^2}} \right)}^{n - 1}}\left( {2\left( 1 \right)} \right)} \right] + {\left( {1 + {1^2}} \right)^n} = 1{}^n{C_0} + 3{}^n{C_1}\left( {{1^2}} \right) + 5{}^n{C_2}\left( {{1^4}} \right) + 7{}^n{C_3}\left( {{1^6}} \right) + .................... + \left( {2n + 1} \right){}^n{C_n}{\left( 1 \right)^{2n}} \\
   \Rightarrow 2n\left( {{2^{n - 1}}} \right) + {2^n} = 1{}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + 7{}^n{C_3} + .................... + \left( {2n + 1} \right){}^n{C_n} \\
   \Rightarrow {2^n}n + {2^n} = 1{}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + 7{}^n{C_3} + .................... + \left( {2n + 1} \right){}^n{C_n} \\
  \therefore 1{}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + 7{}^n{C_3} + .................... + \left( {2n + 1} \right){}^n{C_n} = {2^n}\left( {n + 1} \right) \\
\]
Thus, the correct option is C. \[{2^n}\left( {n + 1} \right)\]

Note: The product rule states that if \[f\left( x \right)\] and \[g\left( x \right)\] are both differentiable, then \[\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]\]. Remember this problem as a formula i.e., \[1{}^n{C_0} + 3{}^n{C_1} + 5{}^n{C_2} + 7{}^n{C_3} + .................... + \left( {2n + 1} \right){}^n{C_n} = {2^n}\left( {n + 1} \right)\] to solve further complicated problems easily.