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Question: If \(S_{n} = \sum_{r = 0}^{n}\frac{1}{n ⥂ C_{r}}\)and \(t_{n} = \sum_{r = 0}^{n}\frac{r}{nC_{r}}\). ...

If Sn=∑r=0n1n⥂CrS_{n} = \sum_{r = 0}^{n}\frac{1}{n ⥂ C_{r}}and tn=∑r=0nrnCrt_{n} = \sum_{r = 0}^{n}\frac{r}{nC_{r}}. Then tnSn\frac{t_{n}}{S_{n}} is equal to

A

2n−12\frac{2n - 1}{2}

B

12n−1\frac{1}{2}n - 1

C

n−1n - 1

D

n2\frac{n}{2}

Answer

n2\frac{n}{2}

Explanation

Solution

Take n=2mn = 2m, then, Sn=12mC0+12mC1+......+12mC2mS_{n} = \frac{1}{2mC_{0}} + \frac{1}{2mC_{1}} + ...... + \frac{1}{2mC_{2m}}

=2[12mC0+12mC1+......+12mCm−1]+12m⥂Cm2\left\lbrack \frac{1}{2mC_{0}} + \frac{1}{2mC_{1}} + ...... + \frac{1}{2mC_{m - 1}} \right\rbrack + \frac{1}{2m ⥂ C_{m}}

tn=∑r=0nrn⥂Cr=∑r=02mr2m⥂Cr=12mC1+22mC2+......+2m2mC2mt_{n} = \sum_{r = 0}^{n}\frac{r}{n ⥂ C_{r}} = \sum_{r = 0}^{2m}\frac{r}{2m ⥂ C_{r}} = \frac{1}{2mC_{1}} + \frac{2}{2mC_{2}} + ...... + \frac{2m}{2mC_{2m}}

tn=(12mC1+2m−12mC2m−1)+(22mC2+2m−22mC2m−2)+......(m−12mCm−1+m+12mCm+1)+m2mCm+2m2mC2mt_{n} = \left( \frac{1}{2mC_{1}} + \frac{2m - 1}{2mC_{2m - 1}} \right) + \left( \frac{2}{2mC_{2}} + \frac{2m - 2}{2mC_{2m - 2}} \right) + ......\left( \frac{m - 1}{2mC_{m - 1}} + \frac{m + 1}{2mC_{m + 1}} \right) + \frac{m}{2mC_{m}} + \frac{2m}{2mC_{2m}}

=2m[12mC1+12mC2+.....]+m2mCm+2m2m\left\lbrack \frac{1}{2mC_{1}} + \frac{1}{2mC_{2}} + ..... \right\rbrack + \frac{m}{2mC_{m}} + 2m

=2m[12mC0+12mC1+...+12mCm−1]+m2mCm=m[Sn−12mCm]+m2mCm=mSn2m\left\lbrack \frac{1}{2mC_{0}} + \frac{1}{2mC_{1}} + ... + \frac{1}{2mC_{m - 1}} \right\rbrack + \frac{m}{2mC_{m}} = m\left\lbrack S_{n} - \frac{1}{2mC_{m}} \right\rbrack + \frac{m}{2mC_{m}} = mSn

tn=mSn⇒tnSn=m=n2t_{n} = mS_{n} \Rightarrow \frac{t_{n}}{S_{n}} = m = \frac{n}{2}