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Binomial Theorem question

2024 · 5 Apr · Shift 1 · Q59
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  5. /2024 · 5 Apr · Shift 1 · Q59

Binomial Theorem question

2024 · 5 Apr · Shift 1 · Q59

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
If the constant term in the expansion of (1+2x−3x3)(32x2−13x)9\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9(1+2x−3x3)(23​x2−3x1​)9 is p\mathrm{p}p, then 108p108 \mathrm{p}108p is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 54

  1. We need the constant term in
(1+2x−3x3)(32x2−13x)9.\left(1+2x-3x^3\right)\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^9.(1+2x−3x3)(23​x2−3x1​)9.

Let

Tr=(9r)(32x2)9−r(−13x)rT_r=\binom{9}{r}\left(\frac{3}{2}x^2\right)^{9-r}\left(-\frac{1}{3x}\right)^rTr​=(r9​)(23​x2)9−r(−3x1​)r

be the general term in the expansion of

(32x2−13x)9.\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^9.(23​x2−3x1​)9.
  1. Simplify the power of xxx in TrT_rTr​:
Tr=(9r)(32)9−r(−13)rx2(9−r)−r=(9r)(32)9−r(−13)rx18−3r.T_r=\binom{9}{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{2(9-r)-r} =\binom{9}{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3r}.Tr​=(r9​)(23​)9−r(−31​)rx2(9−r)−r=(r9​)(23​)9−r(−31​)rx18−3r.

So the exponent of xxx in TrT_rTr​ is

18−3r.18-3r.18−3r.
  1. Now multiply by 1+2x−3x31+2x-3x^31+2x−3x3. For the overall constant term, we need:
  • from 1⋅Tr1\cdot T_r1⋅Tr​: 18−3r=018-3r=018−3r=0
  • from 2x⋅Tr2x\cdot T_r2x⋅Tr​: 18−3r=−118-3r=-118−3r=−1
  • from −3x3⋅Tr-3x^3\cdot T_r−3x3⋅Tr​: 18−3r=−318-3r=-318−3r=−3

Check each:

  • 18−3r=0⇒r=618-3r=0 \Rightarrow r=618−3r=0⇒r=6 (valid)
  • 18−3r=−1⇒3r=1918-3r=-1 \Rightarrow 3r=1918−3r=−1⇒3r=19 (not integer, invalid)
  • 18−3r=−3⇒r=718-3r=-3 \Rightarrow r=718−3r=−3⇒r=7 (valid)

So only two contributions matter:

  • 1⋅T61\cdot T_61⋅T6​
  • −3x3⋅T7-3x^3\cdot T_7−3x3⋅T7​
  1. Compute T6T_6T6​:
T6=(96)(32)3(−13)6x0.T_6=\binom{9}{6}\left(\frac{3}{2}\right)^3\left(-\frac{1}{3}\right)^6 x^0.T6​=(69​)(23​)3(−31​)6x0.

Now,

(96)=84,(32)3=278,(−13)6=1729.\binom{9}{6}=84, \qquad \left(\frac{3}{2}\right)^3=\frac{27}{8}, \qquad \left(-\frac{1}{3}\right)^6=\frac{1}{729}.(69​)=84,(23​)3=827​,(−31​)6=7291​.

Hence

T6=84⋅278⋅1729=84⋅1216=718.T_6=84\cdot \frac{27}{8}\cdot \frac{1}{729} =84\cdot \frac{1}{216} =\frac{7}{18}.T6​=84⋅827​⋅7291​=84⋅2161​=187​.
  1. Compute T7T_7T7​:
T7=(97)(32)2(−13)7x−3.T_7=\binom{9}{7}\left(\frac{3}{2}\right)^2\left(-\frac{1}{3}\right)^7 x^{-3}.T7​=(79​)(23​)2(−31​)7x−3.

Now,

(97)=36,(32)2=94,(−13)7=−12187.\binom{9}{7}=36, \qquad \left(\frac{3}{2}\right)^2=\frac{9}{4}, \qquad \left(-\frac{1}{3}\right)^7=-\frac{1}{2187}.(79​)=36,(23​)2=49​,(−31​)7=−21871​.

Thus

T7=36⋅94⋅(−12187)x−3=81⋅(−12187)x−3=−127x−3.T_7=36\cdot \frac{9}{4}\cdot \left(-\frac{1}{2187}\right)x^{-3} =81\cdot \left(-\frac{1}{2187}\right)x^{-3} =-\frac{1}{27}x^{-3}.T7​=36⋅49​⋅(−21871​)x−3=81⋅(−21871​)x−3=−271​x−3.

Multiplying by −3x3-3x^3−3x3 gives constant contribution

−3x3⋅T7=−3(−127)=19.-3x^3\cdot T_7=-3\left(-\frac{1}{27}\right)=\frac{1}{9}.−3x3⋅T7​=−3(−271​)=91​.
  1. Therefore the constant term ppp is
p=718+19=718+218=918=12.p=\frac{7}{18}+\frac{1}{9}=\frac{7}{18}+\frac{2}{18}=\frac{9}{18}=\frac{1}{2}.p=187​+91​=187​+182​=189​=21​.
  1. Now compute:
108p=108⋅12=54.108p=108\cdot \frac{1}{2}=54.108p=108⋅21​=54.

Therefore the required integer is

54.\boxed{54}.54​.
  1. Comparison with stored answer: Stored correct answer = 545454, which matches our result.
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