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7.Binomial Theorem
hard
यदि ${(1 + x)^m}{(1 - x)^n}$ के प्रसार $(expansion)$ में $x$ और ${x^2}$ के गुणांक $(coefficient)$ क्रमश: $3$ और $-6$ हैं, तो $m =$
A
$6$
B
$9$
C
$12$
D
$24$
(IIT-1999)
Solution
${(1 + x)^m}{(1 – x)^n}$
$ = \left( {1 + mx + \frac{{m(m – 1){x^2}}}{{2!}} + ….} \right)\,\left( {1 – nx + \frac{{n(n – 1)}}{{2!}}{x^2} – ….} \right)$
$ = 1 + (m – n)x + \left[ {\frac{{{n^2} – n}}{2} – mn + \frac{{({m^2} – m)}}{2}} \right]{x^2}$+………
दिया है, $m -n = 3$ या $n = m -3$
अत: $\frac{{{n^2} – n}}{2} – mn + \frac{{{m^2} – m}}{2} = – 6$
$\frac{{(m – 3)(m – 4)}}{2} – m(m – 3) + \frac{{{m^2} – m}}{2} = – 6$
${m^2} – 7m + 12 – 2{m^2} + 6m + {m^2} – m + 12 = 0$
$ – 2m + 24 = 0\,\,\, \Rightarrow m = 12$
Standard 11
Mathematics
