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Chapter 5: Problem 352

If \(a_{n}={ }^{n} \sum_{r=0}\left[1 /\left({ }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}}\right)\right]\), then \(^{\mathrm{n}} \sum_{\mathrm{r}=0}\left[\mathrm{r} /\left({ }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}}\right)\right]\) equals (a) \((\mathrm{n}-1) \mathrm{a}_{\mathrm{n}}\) (b) n \(\mathrm{a}_{\mathrm{n}}\) (c) \((1 / 2) \mathrm{n} \mathrm{a}_{\mathrm{n}}\) (d) None of these

Short Answer

Expert verified
The short answer is: \(n a_{n}\)

Step by step solution

01

Rewrite the summation as a derivative

We will use the fact that the summation of \(a_{n}\) is given as: \[a_{n} = { }^{n} \sum_{r=0} \left[ \frac{1}{ { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}} } \right].\] Now, let's differentiate both sides of this equation with respect to n: \[\frac{d}{dn} a_{n} = \frac{d}{dn} \left( { }^{n} \sum_{r=0} \left[ \frac{1}{ { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}} } \right] \right).\] Differentiating the right side of the equation w.r.t n results in: \[\frac{d}{dn} a_{n} = { }^{n} \sum_{r=0} \left[ \frac{-r}{ { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}}^2 } \frac{d}{dn} \left( { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}} \right) \right].\]
02

Evaluate the derivative of the binomial coefficient

Now, we need to compute the derivative of the binomial coefficient: \[\frac{d}{dn} \left( { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}} \right) = \frac{d}{dn} \left( \frac{n!}{r! (n-r)!} \right).\] Using the property that the derivative of \(n! = \Gamma(n + 1)\) is \(\Gamma'(n + 1)\), we have: \[\frac{d}{dn} \left( { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}} \right) = \frac{\Gamma'(n + 1)}{r! (n-r)!} - \frac{\Gamma'(n - r + 1)}{r! (n-r)!}.\]
03

Substitute the derivative back into the equation

Now, substitute this derivative back into the summation equation: \[\frac{d}{dn} a_{n} = { }^{n} \sum_{r=0} \left[ \frac{-r}{ { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}}^2 } \left( \frac{\Gamma'(n + 1)}{r! (n-r)!} - \frac{\Gamma'(n - r + 1)}{r! (n-r)!} \right) \right].\] Simplifying the equation, we have: \[\frac{d}{dn} a_{n} = { }^{n} \sum_{r=0} \left[ \frac{-r \Gamma'(n + 1)}{ { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}} { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}}} \right],\] which is the summation that we needed to find. Now we need to find which of the given options is equal to the above summation.
04

Compare with the given options

The given options are: (a) \((n-1)a_{n}\) (b) \(n a_{n}\) (c) \((1 / 2)na_{n}\) (d) None of these Since the summation that we have found: \[\frac{d}{dn} a_{n} = { }^{n} \sum_{r=0} \left[ \frac{-r \Gamma'(n + 1)}{ { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}} { }^{\mathrm{r}} \mathrm{C}_{\mathrm{n}}} \right],\] it resembles the form \(\frac{d}{dn} a_{n} = n a_{n} \). Hence, the correct answer is: (b) \(n a_{n}\).

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Most popular questions from this chapter

How many number greater than 10 lac be formed from \(2,3,0,3,4,2,3\) (a) 420 (b) 360 (c) 400 (d) 300

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