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The two consecutive terms in the expansi...

The two consecutive terms in the expansion of `(1+x)^(24)` whose coefficients are in the ratio 1:4 are

A

(a) 3rd and 4th

B

(b) 4th and 5th

C

(c) 5th and 6th

D

(d) 6th and 7th

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The correct Answer is:
To solve the problem of finding the two consecutive terms in the expansion of \((1+x)^{24}\) whose coefficients are in the ratio \(1:4\), we can follow these steps: ### Step 1: Identify the general term The general term \(T_{r+1}\) in the expansion of \((1+x)^n\) is given by the formula: \[ T_{r+1} = \binom{n}{r} A^{n-r} B^r \] For our case, \(A = 1\), \(B = x\), and \(n = 24\). Thus, the general term becomes: \[ T_{r+1} = \binom{24}{r} x^r \] ### Step 2: Write the consecutive terms The two consecutive terms we are interested in are \(T_{r+1}\) and \(T_{r+2}\): - The \(r+1\)th term is: \[ T_{r+1} = \binom{24}{r} x^r \] - The \(r+2\)th term is: \[ T_{r+2} = \binom{24}{r+1} x^{r+1} \] ### Step 3: Set up the ratio of coefficients According to the problem, the coefficients of these terms are in the ratio \(1:4\). Therefore, we can write: \[ \frac{\binom{24}{r}}{\binom{24}{r+1}} = \frac{1}{4} \] ### Step 4: Simplify the ratio Using the property of binomial coefficients: \[ \frac{\binom{n}{r}}{\binom{n}{r+1}} = \frac{r+1}{n-r} \] we can substitute \(n = 24\): \[ \frac{\binom{24}{r}}{\binom{24}{r+1}} = \frac{r+1}{24 - r} \] Thus, we have: \[ \frac{r+1}{24 - r} = \frac{1}{4} \] ### Step 5: Cross-multiply and solve for \(r\) Cross-multiplying gives: \[ 4(r + 1) = 1(24 - r) \] Expanding both sides: \[ 4r + 4 = 24 - r \] Combining like terms: \[ 4r + r = 24 - 4 \] \[ 5r = 20 \] Dividing both sides by 5: \[ r = 4 \] ### Step 6: Identify the terms Now that we have \(r = 4\), we can find the two consecutive terms: - The \(r+1\)th term (5th term) is: \[ T_5 = \binom{24}{4} x^4 \] - The \(r+2\)th term (6th term) is: \[ T_6 = \binom{24}{5} x^5 \] ### Conclusion Thus, the two consecutive terms in the expansion of \((1+x)^{24}\) whose coefficients are in the ratio \(1:4\) are the 5th and 6th terms. ### Final Answer The answer is option C: 5th and 6th. ---
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