Question
Download Solution PDFयदि (1 + x)m के द्विपद प्रसरण में तीसरा पद (-1/8)x² है, तो m का परिमेय मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
(1 + x)n 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 = \(\frac 12\)
Last updated on Jun 11, 2025
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