Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If we have the coefficients ${{C}_{0}},{{C}_{1}},{{C}_{2}},...,{{C}_{n}}$ are binomial coefficients in the expansion of ${{\left( 1+x \right)}^{n}}$ the value of ${{C}_{0}}+\dfrac{{{C}_{1}}}{2}+\dfrac{{{C}_{2}}}{3}+...+\dfrac{{{C}_{n}}}{n+1}$ equals?

seo-qna
Last updated date: 16th Jul 2024
Total views: 348.3k
Views today: 3.48k
Answer
VerifiedVerified
348.3k+ views
Hint: In the given question we need to use the concept of binomial expansion and then also make use of the definition of combination as in the question we are been asked the sum of the coefficients of the binomial expansion of the given term ${{\left( 1+x \right)}^{n}}$. And the sum of coefficients involves n+1 coefficients of the given expansion.

Complete step-by-step solution:
According to the given question, we need to find the sum of the coefficients of the n+1 terms of the expansion. Also, we need to find the sum of ${{C}_{0}}+\dfrac{{{C}_{1}}}{2}+\dfrac{{{C}_{2}}}{3}+...+\dfrac{{{C}_{n}}}{n+1}$.
So, we can also write this as $\sum\limits_{r=0}^{r=n}{\dfrac{{{C}_{r}}}{r+1}}$ .
Now, we know that the $r^{th}$ coefficient of binomial expansion is given by ${}^{n}C_{r}=\dfrac{n!}{r!\left( n-r \right)!}$ .
Now, using the above two expression we get:
 $\sum\limits_{r=0}^{r=n}{\dfrac{{{C}_{r}}}{r+1}}=\dfrac{n!}{r!\left( n-r \right)!}\times \dfrac{1}{r+1}$
Now, multiplying and dividing by n+1 the above term we get, $\begin{align}
  & \sum\limits_{r=0}^{r=n}{\dfrac{{{C}_{r}}}{r+1}}=\dfrac{\left( n+1 \right)n!}{r!\left( n-r \right)!}\times \dfrac{1}{r+1}\times \dfrac{1}{\left( n+1 \right)} \\
 &= \dfrac{1}{\left( n+1 \right)}\sum\limits_{r=0}^{r=n}{\dfrac{\left( n+1 \right)!}{\left( r+1 \right)!}\times \dfrac{1}{\left( n-r \right)!}} \\
 &= \dfrac{1}{\left( n+1 \right)}\sum\limits_{r=0}^{r=n}{\dfrac{\left( n+1 \right)!}{\left( r+1 \right)!}\times \dfrac{1}{\left( n+1-\left( r+1 \right) \right)!}} \\
 &= \dfrac{1}{\left( n+1 \right)}\sum\limits_{r=0}^{r=n+1}{C_{r+1}^{n+1}} \\
\end{align}$
Now, we know that the expansion of ${{\left( 1+x \right)}^{n}}$is ${{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+...$
Now, when we take x=1, we get ${}^{n}C_{0}+{}^{n}C_{1}+....+{}^{n}C_{r}+...$
Now similarly if we take n+1 term we get, ${}^{n+1}C_{0}+{}^{n+1}C_{1}+....+{}^{n+1}C_{n+1}$ , now using the above expression we get
$\begin{align}
  &= \dfrac{1}{\left( n+1 \right)}\sum\limits_{r=0}^{r=n}{{}^{n+1}C_{r+1}}-\dfrac{1}{\left( n+1 \right)}{}^{n+1}C_{0}+\dfrac{1}{\left( n+1 \right)}{}^{n+1}C_{0} \\
&= \dfrac{1}{\left( n+1 \right)}\sum\limits_{r=0}^{r=n}{{}^{n+1}C_{r}}-\dfrac{1}{\left( n+1 \right)}{}^{n+1}C_{0} \\
 & = \dfrac{{{2}^{n+1}}-1}{n+1} \\
\end{align}$
Therefore, the sum of the coefficients of the binomial expansion of the term ${{\left( 1+x \right)}^{n}}$is $\dfrac{{{2}^{n+1}}-1}{n+1}$.

Note: In these types of questions, we need to take care of the calculation initially. Now, we need to be very careful with the concept of the combination, then only we will be able to find out the way in which we can get the sum of the given expansion. Such questions are although easy to solve but we get confused in the concept itself as these questions are not that direct.