Question:
In the expansion of \((1 + ax)^n\), the first three terms are respectively 1, \(12x\), and \(64x^2\). What is \(n\) equal to?
\((1 + ax)^n\) के विस्तार में प्रथम तीन पद क्रमशः 1, \(12x\), और \(64x^2\) हैं। \(n\) किसके बराबर है?
Options:
- 6
- 9
- 10
- 12
- 6
- 9
- 10
- 12
Correct Answer: b)
Explanation:
The expansion of \((1 + ax)^n\) begins with:
\[(1 + ax)^n = 1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \dots = 1 + nax + \frac{n(n-1)a^2}{2}x^2 + \dots\]
Equating coefficients:
1. \(na = 12 \implies a = \frac{12}{n}\)
2. \(\frac{n(n-1)a^2}{2} = 64 \implies n(n-1)a^2 = 128\)
Substitute \(a = \frac{12}{n}\):
\[n(n-1)\left(\frac{144}{n^2}\right) = 128 \implies \frac{n-1}{n} = \frac{128}{144} = \frac{8}{9}\]
\[9(n – 1) = 8n \implies 9n – 9 = 8n \implies n = 9\]
Hence, the correct option is (b).
\((1 + ax)^n\) का प्रसार:
\(1 + nax + \frac{n(n-1)}{2}a^2 x^2 + \dots\)।
तुलना करने पर:
1. \(na = 12 \implies a = 12/n\)
2. \(\frac{n(n-1)}{2}a^2 = 64\)
मान रखने पर:
\[\frac{n-1}{n} \times 144 = 128 \implies \frac{n-1}{n} = \frac{8}{9} \implies n = 9\]
अतः सही विकल्प (b) है।