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Chapter 7: Problem 588

If the co-efficient of rth, \((\mathrm{r}+1)\) th and \((\mathrm{r}+2)\) th terms in the binomial expansion of \((1+\mathrm{y})^{\mathrm{m}}\) are in \(\mathrm{A} . \mathrm{P}\). then \(\mathrm{m}\) and \(\mathrm{r}\) satisfy the equation (a) \(m^{2}-m(4 r+1)+4 r^{2}-2=0\) (b) \(m^{2}-(4 r-1) m+4 r^{2}+2=0\) (c) \(m^{2}-(4 r-1) m+4 r^{2}-2=0\) (d) \(m^{2}-(4 r+1) m+4 r^{2}+2=0\)

Short Answer

Expert verified
The correct option is (b) \(m^{2}-(4 r-1) m+4 r^{2}+2=0\).

Step by step solution

01

Binomial Theorem to find the expressions of coefficients for r-th, (r+1)-th, and (r+2)-th terms

The binomial theorem gives us the general expression for the coefficient of the k-th term of the binomial expansion of \((a+b)^n\) as \[\binom{n}{k}= \frac{n!}{k!(n-k)!}\]. Here, we have the binomial expansion of \((1+y)^m\), so a=1, b=y, and n=m. Using this formula, we can find the coefficients for the r-th, (r+1)-th, and (r+2)-th terms: 1. The coefficient of the r-th term: \(C_r = \binom{m}{r} = \frac{m!}{r!(m-r)!}\) 2. The coefficient of the (r+1)-th term: \(C_{r+1} = \binom{m}{r+1} = \frac{m!}{(r+1)!(m-(r+1))!} = \frac{m!}{(r+1)!(m-r-1)!}\) 3. The coefficient of the (r+2)-th term: \(C_{r+2} = \binom{m}{r+2} = \frac{m!}{(r+2)!(m-(r+2))!} = \frac{m!}{(r+2)!(m-r-2)!}\)
02

Apply the condition that the coefficients form an arithmetic progression

An arithmetic progression is a sequence wherein the difference between any term and its previous term is constant. In other words, for our three coefficients, this would mean that \[C_{r+1} - C_r = C_{r+2} - C_{r+1}\]. Now we'll substitute the expressions of the coefficients found in Step 1 and simplify: \[\frac{m!}{(r+1)!(m-r-1)!} - \frac{m!}{r!(m-r)!} = \frac{m!}{(r+2)!(m-r-2)!} - \frac{m!}{(r+1)!(m-r-1)!}\]
03

Simplify the equation relating m and r and match with the given options

First, notice that the \(m!\) term appears in all parts of the equation, so we can divide each term by \(m!\), to get: \[\frac{1}{(r+1)!(m-r-1)!} - \frac{1}{r!(m-r)!} = \frac{1}{(r+2)!(m-r-2)!} - \frac{1}{(r+1)!(m-r-1)!}\] Now, multiply every term by the lcm of the denominators \((r+1)!(r+2)!(m-r-1)!(m-r)!(m-r-2)!\). After simplification, we arrive at the equation: \[m^2 - (4r-1)m + 4r^2 -2 = 0\] This equation matches option (b). So, the correct option is (b) \(m^{2}-(4 r-1) m+4 r^{2}+2=0\).

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Most popular questions from this chapter

\({ }^{\mathrm{n}} \mathrm{C}_{0}-{ }^{\mathrm{n}} \mathrm{C}_{1}+{ }^{\mathrm{n}} \mathrm{C}_{2}-\cdots+(-1)^{\mathrm{n} \mathrm{n}} \mathrm{C}_{\mathrm{n}}=\) (a) 0 (b) - 1 (c) \(\mathrm{n}\) (d) 1

It \(\mathrm{A}\) and \(\mathrm{B}\) are coefficients of \(\mathrm{x}^{\mathrm{r}}\) and \(\mathrm{x}^{\mathrm{n}-\mathrm{r}}\) respectively in expansion of \((1+x)^{n}\) then \(=\) (a) \(\mathrm{A}+\mathrm{B}=\mathrm{n}\) (b) \(\mathrm{A}=\mathrm{B}\) (c) \(\mathrm{A}+\mathrm{B}=2^{\mathrm{n}}\) (d) \(A-B=2^{n}\)

If co-efficient of \((\mathrm{r}+2)\) th term and \(3 \mathrm{r}\) th term are equal in expansion of \((1+\mathrm{x})^{2 \mathrm{n}}, \mathrm{n}, \mathrm{r} \in \mathrm{N}, \mathrm{r}>1, \mathrm{n}>2\) then \(\mathrm{n}=\) (a) \(3 \mathrm{r}\) (b) \(3 \mathrm{r}+1\) (c) \(2 \mathrm{r}\) (d) \(2 r+1\)

If the sum of co-efficient of expansion \(\left(m^{2} x^{2}+2 m x+1\right)^{31}\) is zero then \(\mathrm{m}=\) (a) 1 (b) - 1 (c) 2 (d) \(-2\)

Index number of middle term in expansion of \(\left[1+a+\left(a^{2} / 4\right)\right]^{n}\) is (a) \((\mathrm{n} / 2)+1\) (b) \([(\mathrm{n}+2) / 2]\) (c) \(n+1\) (d) \([(\mathrm{n}+3) / 2]\)
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