JEE Mains · Maths · STD 11 - 7. binomial theoram
If the constant term in the expansion of \(\left(3 x^{3}-2 x^{2}+\frac{5}{x^{5}}\right)^{10}\) is \(2^{k} . l\), where \(l\) is an odd integer, then the value of \(k\) is equal to
- A \(6\)
- B \(7\)
- C \(8\)
- D \(9\)
Answer & Solution
Correct Answer
(D) \(9\)
Step-by-step Solution
Detailed explanation
General term \(T _{ r +1}=\frac{! 10}{! r _{1}! r _{2}! r _{3}}(3)^{ r _{1}}(-2)^{ r _{2}}(5)^{ r _{3}}( x )^{3 r _{1}+2 r _{2}-5 r _{3}}\) \(3 r_{1}+2 r_{2}-5 r_{3}=0\) \(\dots(1)\) \(r_{1}+r_{2}+r_{3}=10\) \(\dots(2)\) from equation \((1)\) and \((2)\)…
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