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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2015 · Mathematics · Sequences and Series

If binomial coefficients of three consecutive terms of (1+x)^ 2 are...

If binomial coefficients of three consecutive terms of (1+x)2(1+x)^{2} are in HPH P, then the maximum value of n\mathrm{n} is

Options

  1. A1
  2. B2
  3. C0
  4. DNone of these

Correct answer

D. None of these

Step-by-step solution

Let the coefficients of rth, (r+1)(r+1) th, and (r +2)thtermsbeinHP.Then,2(nr)=1(nr−1)+1(nr)+1+ 2) th terms be in HP. Then, \frac{2}{\binom{n}{r}}=\frac{1}{\binom{n}{r-1}}+\frac{1}{\binom{n}{r}+1}⇒2=nCrnCr−1=nCrnCr+1⇒2=n−r+1r+r+1n−r⇒\mathrmn2−4nr+4\mathrmr2+n=0⇒(n−2r)2+n=0\begin{array}{l} \Rightarrow \quad2=\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}}\\ \Rightarrow \quad 2=\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}+\frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}}\\ \Rightarrow \mathrmn^{2}-4 \mathrm{nr}+4 \mathrmr^{2}+\mathrm{n}=0 \\ \Rightarrow(\mathrm{n}-2 \mathrm{r})^{2}+\mathrm{n}=0 \end{array}whichisnotpossibleforanyvalueforwhich is not possible for any value for n..
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