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Question: If \({{a}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{{}^{n}{{C}_{r}}}}\) then \(\sum\limits_{r=0}^{n}{\dfr...

If an=∑r=0n1nCr{{a}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{{}^{n}{{C}_{r}}}} then ∑r=0nrnCr\sum\limits_{r=0}^{n}{\dfrac{r}{{}^{n}{{C}_{r}}}} equals?
1. (n−1)an\left( n-1 \right){{a}_{n}}
2. nann{{a}_{n}}
3. n2an\dfrac{n}{2}{{a}_{n}}
4. None of these

Explanation

Solution

For solving this question you should know about the general formulas of combinations. As we know that nCr{}^{n}{{C}_{r}} can also be written as nCn−r{}^{n}{{C}_{n-r}} easily. So, we will use this relation here. Here, we will first assume another value as rnCr\dfrac{r}{{}^{n}{{C}_{r}}} and then we will expand and solve it further. And by solving this we will get the required final answer.

Complete step-by-step solution:
According to the question it is asked about the value of ∑r=0nrnCr\sum\limits_{r=0}^{n}{\dfrac{r}{{}^{n}{{C}_{r}}}} and it is given that an=∑r=0n1nCr{{a}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{{}^{n}{{C}_{r}}}}.
Now let us assume that,
b=∑r=0nrnCrb=\sum\limits_{r=0}^{n}{\dfrac{r}{{}^{n}{{C}_{r}}}}
And this can also be written as,
b=∑r=0nn−(n−r)nCrb=\sum\limits_{r=0}^{n}{\dfrac{n-\left( n-r \right)}{{}^{n}{{C}_{r}}}}
And now if we solve this, then we will get as follows,
b=∑r=0nnnCr−∑r=0n(n−r)nCrb=\sum\limits_{r=0}^{n}{\dfrac{n}{{}^{n}{{C}_{r}}}}-\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{{}^{n}{{C}_{r}}}}
We will now further take nn outside and make it as a term of an{{a}_{n}}. So, by doing that we will get the new values as follows,
b=n∑r=0n1nCr−∑r=0n(n−r)nCrb=n\sum\limits_{r=0}^{n}{\dfrac{1}{{}^{n}{{C}_{r}}}}-\sum\limits_{r=0}^{n}{\dfrac{\left( n-r \right)}{{}^{n}{{C}_{r}}}}
Since it is already given in the question that an=∑r=0n1nCr{{a}_{n}}=\sum\limits_{r=0}^{n}{\dfrac{1}{{}^{n}{{C}_{r}}}}, so, we will use the value in the above expression and then we will get as follows,
b=nan−∑r=0nn−rnCrb=n{{a}_{n}}-\sum\limits_{r=0}^{n}{\dfrac{n-r}{{}^{n}{{C}_{r}}}}
Therefore we have,
nCr=nCn−r ⇒b=nan−∑r=0nn−rnCr \begin{aligned} & {}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}} \\\ & \Rightarrow b=n{{a}_{n}}-\sum\limits_{r=0}^{n}{\dfrac{n-r}{{}^{n}{{C}_{r}}}} \\\ \end{aligned}
Solving it further we will get as follows,
b=nan−b ⇒2b=nan ⇒b=n2an \begin{aligned} & b=n{{a}_{n}}-b \\\ & \Rightarrow 2b=n{{a}_{n}} \\\ & \Rightarrow b=\dfrac{n}{2}{{a}_{n}} \\\ \end{aligned}
Hence, we get the final value of b=n2anb=\dfrac{n}{2}{{a}_{n}} and so the correct answer is option.

Note: While solving these types of questions you have to be careful about the given values. And we have to add the values as it can be again with the same terms. If we take other values, then it will expand very much and there will be a lot of calculations and it will lead to less chance of a correct answer.