यदि (1 + x)m के द्विपद प्रसरण में तीसरा पद (-1/8)x² है, तो m का परिमेय मान क्या है?

  1. 2
  2. \(\frac 12\)
  3. 3
  4. इनमें से कोई नहीं 

Answer (Detailed Solution Below)

Option 2 : \(\frac 12\)
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Detailed Solution

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संकल्पना:

(1 + x)का प्रसरण:

\(\rm (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3 +....\)

गणना:

दिया गया है: (1 + x)m के द्विपद प्रसरण में तीसरा पद (-1/8)x² है। 

\(\rm (1+x)^m= 1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3 +....\)

इसलिए, (1 + x)m के द्विपद प्रसरण में तीसरा पद\(\rm \frac{m(m-1)}{2!}x^2\) है। 

\(\rm \frac{m(m-1)}{2!}x^2\) = (-1/8)x2

⇒ \(\rm \frac{m(m-1)}{2}= \frac {-1}{8}\)

⇒ 4m2 - 4m + 1 = 0

⇒ (2m - 1)2 = 0

⇒ 2m - 1 = 0

∴ m = \(\frac 12\)

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