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Binomial Theorem question

2023 · 13 Apr · Shift 2 · Q27
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  5. /2023 · 13 Apr · Shift 2 · Q27

Binomial Theorem question

2023 · 13 Apr · Shift 2 · Q27

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
The coefficient of x5x^{5}x5 in the expansion of (2x3−13x2)5\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}(2x3−3x21​)5 is :
  1. A
    263\frac{26}{3}326​
  2. B
    809\frac{80}{9}980​
  3. C
    9
  4. D
    8
View written solutionFree

Correct answer: B

  1. Consider the binomial expansion
(2x3−13x2)5.\left(2x^3-\frac{1}{3x^2}\right)^5.(2x3−3x21​)5.

Its general term is

Tr+1=(5r)(2x3)5−r(−13x2)r,T_{r+1}=\binom{5}{r}(2x^3)^{5-r}\left(-\frac{1}{3x^2}\right)^r,Tr+1​=(r5​)(2x3)5−r(−3x21​)r,

where r=0,1,2,3,4,5r=0,1,2,3,4,5r=0,1,2,3,4,5.

  1. Simplify the power of xxx in the general term:
(2x3)5−r=25−rx3(5−r)=25−rx15−3r,(2x^3)^{5-r}=2^{5-r}x^{3(5-r)}=2^{5-r}x^{15-3r},(2x3)5−r=25−rx3(5−r)=25−rx15−3r,

and

(−13x2)r=(−1)r3rx−2r.\left(-\frac{1}{3x^2}\right)^r = \frac{(-1)^r}{3^r}x^{-2r}.(−3x21​)r=3r(−1)r​x−2r.

So,

Tr+1=(5r)25−r(−1)r3rx15−3r−2r=(5r)25−r(−1)r3rx15−5r.T_{r+1}=\binom{5}{r}2^{5-r}\frac{(-1)^r}{3^r}x^{15-3r-2r} =\binom{5}{r}2^{5-r}\frac{(-1)^r}{3^r}x^{15-5r}.Tr+1​=(r5​)25−r3r(−1)r​x15−3r−2r=(r5​)25−r3r(−1)r​x15−5r.
  1. We need the coefficient of x5x^5x5, so set
15−5r=5.15-5r=5.15−5r=5.

Thus,

5r=10  ⟹  r=2.5r=10 \implies r=2.5r=10⟹r=2.
  1. Substitute r=2r=2r=2 into the coefficient:
(52)25−2(−1)232=10⋅23⋅19=10⋅8⋅19=809.\binom{5}{2}2^{5-2}\frac{(-1)^2}{3^2} =10\cdot 2^3\cdot \frac{1}{9} =10\cdot 8\cdot \frac{1}{9} =\frac{80}{9}.(25​)25−232(−1)2​=10⋅23⋅91​=10⋅8⋅91​=980​.
  1. Therefore, the coefficient of x5x^5x5 is
809.\boxed{\frac{80}{9}}.980​​.
  1. Option check:
  • A: 263\frac{26}{3}326​ ❌
  • B: 809\frac{80}{9}980​ ✅
  • C: 999 ❌
  • D: 888 ❌

Hence, the correct option is B.

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