Question

Evaluate the sum of the series ${}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+.....+{}^{n}{{C}_{n}}$ equals to
(a)${{2}^{n}}-1$
(b)${{2}^{n}}$
(c)${{2}^{n}}+1$
(d)None of these

Answer 1

Hint : Firstly, we will use the general expansion of the term ${{\left( 1+x \right)}^{n}}$by using the binomial expansion as ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{0}}+{}^{n}{{C}_{1}}{{x}^{1}}+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{3}}{{x}^{3}}+.....+{}^{n}{{C}_{n}}{{x}^{n}}$. Then, we will substitute x=1 to get the question similar to the expression. Then, we will find the value of the expression ${}^{n}{{C}_{0}}$ by using the formula of the combination as $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Then, we get the final answer after rearranging the equation.

Complete step by step solution : In this question, we will use the general expansion of the term ${{\left( 1+x \right)}^{n}}$by using the binomial expansion which is given by:
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{0}}+{}^{n}{{C}_{1}}{{x}^{1}}+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{3}}{{x}^{3}}+.....+{}^{n}{{C}_{n}}{{x}^{n}}$
To make the above expression similar to the question, substitute x=1 in the above expression.
By substituting x=1, we get
$\begin{aligned} & {{\left( 1+1 \right)}^{n}}={}^{n}{{C}_{0}}{{\left( 1 \right)}^{0}}+{}^{n}{{C}_{1}}{{\left( 1 \right)}^{1}}+{}^{n}{{C}_{2}}{{\left( 1 \right)}^{2}}+{}^{n}{{C}_{3}}{{\left( 1 \right)}^{3}}+.....+{}^{n}{{C}_{n}}{{\left( 1 \right)}^{n}} \\\ & \Rightarrow {{2}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+.....+{}^{n}{{C}_{n}}.......\left( i \right) \\\ \end{aligned}$
Now, to get the value of ${}^{n}{{C}_{0}}$ is given by the formula of the combination:
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
To find the factorial of the number n, multiply the number n with (n-1) till it reaches to 1. To understand let us find the factorial of 4.
$\begin{aligned} & 4!=4\times 3\times 2\times 1 \\\ & \Rightarrow 24 \\\ \end{aligned}$
Now, we substitute r=0 in the above formula and get the value of ${}^{n}{{C}_{0}}$:
$\begin{aligned} & {}^{n}{{C}_{0}}=\dfrac{n!}{\left( n-0 \right)!0!} \\\ & \Rightarrow \dfrac{n!}{n!} \\\ & \Rightarrow 1 \\\ \end{aligned}$
Substituting the value of ${}^{n}{{C}_{0}}$ in equation (i), we get:
${{2}^{n}}=1+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+.....+{}^{n}{{C}_{n}}$
Now subtracting 1 from the sides of the above equation, we get:
$\begin{aligned} & {{2}^{n}}-1=1+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+.....+{}^{n}{{C}_{n}}-1 \\\ & \Rightarrow {{2}^{n}}-1={}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+.....+{}^{n}{{C}_{n}}....\left( ii \right) \\\ \end{aligned}$
So, equation (ii) gives the final sum of the required series.
Hence, option (a) is correct.

Note : Here, we can make mistakes in taking the value of the $0!$ as most of us take is zero which will lead to the value of ${}^{n}{{C}_{0}}$ as infinity. But, $0!$is the special case of the factorial that is equal to 1 only. So, instead of taking it as 0 consider ${}^{n}{{C}_{0}}$ as 1 and solve the remaining question accordingly.