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Question: If \({{\text{t}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{1}{{{{({}^{\...

If tn = ∑r = 0n1(nCr)k{{\text{t}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{1}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} and Sn = ∑r = 0nr(nCr)k{{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{\text{r}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} , where k∈Z+{\text{k}} \in {{\text{Z}}^ + }, then cos⁡−1(Snntn){\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{n}}{{\text{t}}_{\text{n}}}}}} \right)
A. π6\dfrac{\pi }{6}
B. π4\dfrac{\pi }{4}
C. π3\dfrac{\pi }{3}
D. π2\dfrac{\pi }{2}

Explanation

Solution

- Hint: To solve this problem, we will use the property nCr = nCn - r{}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}{}^{\text{n}}{{\text{C}}_{{\text{n - r}}}}. We will simplify tn{{\text{t}}_{\text{n}}} and Sn{{\text{S}}_{\text{n}}} by using this property and also find a relation between them. Then we will use the trigonometric value of the inverse function cos−1x{\text{co}}{{\text{s}}^{ - 1}}{\text{x}} to find the value of cos⁡−1(Snntn){\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{n}}{{\text{t}}_{\text{n}}}}}} \right).

Complete step-by-step solution -

Now, we are given
tn = ∑r = 0n1(nCr)k{{\text{t}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{1}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} and Sn = ∑r = 0nr(nCr)k{{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{\text{r}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}}
Now, r can be written as r = n – (n – r), Therefore, Sn{{\text{S}}_{\text{n}}} can be written as
Sn = ∑r = 0nn - (n - r)(nCr)k{{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - (n - r)}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}}
Sn = ∑r = 0nn (nCr)k - ∑r = 0nn - r(nCr)k{{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n }}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} {\text{ - }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}}
Sn = n∑r = 0n1 (nCr)k - ∑r = 0nn - r(nCr)k{{\text{S}}_{\text{n}}}{\text{ = n}}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{1 }}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} {\text{ - }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}}
Therefore, Sn = ntn - ∑r = 0nn - r(nCr)k{{\text{S}}_{\text{n}}}{\text{ = n}}{{\text{t}}_{\text{n}}}{\text{ - }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} … (1)
Now, let bn = ∑r = 0nn - r(nCr)k{{\text{b}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}}
From the property nCr = nCn - r{}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}{}^{\text{n}}{{\text{C}}_{{\text{n - r}}}}, we get
bn = ∑r = 0nn - r(nCn - r)k{{\text{b}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{{\text{n - r}}}}{{{{({}^{\text{n}}{{\text{C}}_{{\text{n - r}}}})}^{\text{k}}}}}}
Expanding bn{{\text{b}}_{\text{n}}}, we get
bn = n(nCn)k + n - 1(nCn - 1)k + n - 2(nCn - 2)k + ........ + 1(nC1)k + 0(nC0)k{{\text{b}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{n}}})}^{\text{k}}}}}{\text{ + }}\dfrac{{{\text{n - 1}}}}{{{{({}^{\text{n}}{{\text{C}}_{{\text{n - 1}}}})}^{\text{k}}}}}{\text{ + }}\dfrac{{{\text{n - 2}}}}{{{{({}^{\text{n}}{{\text{C}}_{{\text{n - }}2}})}^{\text{k}}}}}{\text{ + }}........{\text{ + }}\dfrac{{\text{1}}}{{{{({}^{\text{n}}{{\text{C}}_1})}^{\text{k}}}}}{\text{ + }}\dfrac{0}{{{{({}^{\text{n}}{{\text{C}}_0})}^{\text{k}}}}}
Also, Sn = ∑r = 0nr(nCr)k{{\text{S}}_{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^{\text{n}} {\dfrac{{\text{r}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{r}}})}^{\text{k}}}}}} .So, expanding Sn{{\text{S}}_{\text{n}}}, we get
Sn = 0(nC0)k + 1(nC1)k + 2(nC2)k + ........ + n - 1(nCn - 1)k + n(nCn)k{{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{0}{{{{({}^{\text{n}}{{\text{C}}_0})}^{\text{k}}}}}{\text{ + }}\dfrac{1}{{{{({}^{\text{n}}{{\text{C}}_1})}^{\text{k}}}}}{\text{ + }}\dfrac{2}{{{{({}^{\text{n}}{{\text{C}}_2})}^{\text{k}}}}}{\text{ + }}........{\text{ + }}\dfrac{{{\text{n - 1}}}}{{{{({}^{\text{n}}{{\text{C}}_{{\text{n - 1}}}})}^{\text{k}}}}}{\text{ + }}\dfrac{{\text{n}}}{{{{({}^{\text{n}}{{\text{C}}_{\text{n}}})}^{\text{k}}}}}
Now, we can see that both the expansion of bn{{\text{b}}_{\text{n}}} and Sn{{\text{S}}_{\text{n}}} are equal. Therefore, we get
Sn{{\text{S}}_{\text{n}}} = bn{{\text{b}}_{\text{n}}}. So, from equation (1), we get
Sn = ntn - Sn{{\text{S}}_{\text{n}}}{\text{ = n}}{{\text{t}}_{\text{n}}}{\text{ - }}{{\text{S}}_{\text{n}}}
2Sn = ntn{\text{2}}{{\text{S}}_{\text{n}}}{\text{ = n}}{{\text{t}}_{\text{n}}}
Now, cos⁡−1(Snntn){\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{n}}{{\text{t}}_{\text{n}}}}}} \right) = cos⁡−1(Sn2Sn){\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{2}}{{\text{S}}_{\text{n}}}}}} \right) = cos⁡−1(12){\cos ^{ - 1}}\left( {\dfrac{1}{{\text{2}}}} \right). Now cos⁡−1(12){\cos ^{ - 1}}\left( {\dfrac{1}{{\text{2}}}} \right) = π3\dfrac{\pi }{3}
Therefore, cos⁡−1(Snntn){\cos ^{ - 1}}\left( {\dfrac{{{{\text{S}}_{\text{n}}}}}{{{\text{n}}{{\text{t}}_{\text{n}}}}}} \right) = π3\dfrac{\pi }{3}
So, option (C) is correct.

Note: When we come up with such types of questions, we have to first simplify the given terms. Then, we have to find a relation between given terms as the question can be solved easily with the help of the relation. We have to use the property nCr = nCn - r{}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}{}^{\text{n}}{{\text{C}}_{{\text{n - r}}}} because it plays the most important role in finding the relation and solving the given problem.