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Binomial Theorem question

2022 · 27 Jul · Shift 2 · Q35
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  5. /2022 · 27 Jul · Shift 2 · Q35

Binomial Theorem question

2022 · 27 Jul · Shift 2 · Q35

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
Let for the 9th 9^{\text {th }}9th  term in the binomial expansion of (3+6x)n(3+6 x)^{\mathrm{n}}(3+6x)n, in the increasing powers of 6x6 x6x, to be the greatest for x=32x=\frac{3}{2}x=23​, the least value of n\mathrm{n}n is n0\mathrm{n}_{0}n0​. If k\mathrm{k}k is the ratio of the coefficient of x6x^{6}x6 to the coefficient of x3x^{3}x3, then k+n0\mathrm{k}+\mathrm{n}_{0}k+n0​ is equal to :
Numerical answer
View written solutionFree

Correct answer: 24

  1. General term of the expansion

For

(3+6x)n,(3+6x)^n,(3+6x)n,

the general term in increasing powers of 6x6x6x is

Tr+1=(nr)3n−r(6x)r.T_{r+1}=\binom{n}{r}3^{n-r}(6x)^r.Tr+1​=(rn​)3n−r(6x)r.

So the 9th9^{\text{th}}9th term corresponds to r=8r=8r=8:

T9=(n8)3n−8(6x)8.T_9=\binom{n}{8}3^{n-8}(6x)^8.T9​=(8n​)3n−8(6x)8.
  1. Condition for the 9th term to be the greatest

At

x=32,x=\frac{3}{2},x=23​,

we get

6x=6⋅32=9.6x=6\cdot \frac{3}{2}=9.6x=6⋅23​=9.

Thus the expansion becomes effectively

(3+9)n,(3+9)^n,(3+9)n,

and the magnitude of the (r+1)(r+1)(r+1)-th term is proportional to

(nr)3n−r9r.\binom{n}{r}3^{n-r}9^r.(rn​)3n−r9r.

For the 9th9^{\text{th}}9th term to be the greatest, we need:

  • T9≥T8T_9\ge T_8T9​≥T8​
  • T9≥T10T_9\ge T_{10}T9​≥T10​

Using

Tr+1Tr=n−r+1r⋅93=3⋅n−r+1r,\frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}\cdot \frac{9}{3}=3\cdot \frac{n-r+1}{r},Tr​Tr+1​​=rn−r+1​⋅39​=3⋅rn−r+1​,

more directly,

\frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}\cdot \frac{6x}{3}= rac{n-r+1}{r}\cdot 2x.

At x=32x=\frac32x=23​, this becomes

Tr+1Tr=3⋅n−r+1r.\frac{T_{r+1}}{T_r}=3\cdot \frac{n-r+1}{r}.Tr​Tr+1​​=3⋅rn−r+1​.

Now:

Compare T9T_9T9​ and T8T_8T8​

Here r=8r=8r=8 for T9/T8T_9/T_8T9​/T8​:

T9T8=3⋅n−8+18=3⋅n−78.\frac{T_9}{T_8}=3\cdot \frac{n-8+1}{8}=3\cdot \frac{n-7}{8}.T8​T9​​=3⋅8n−8+1​=3⋅8n−7​.

For T9≥T8T_9\ge T_8T9​≥T8​,

3⋅n−78≥13\cdot \frac{n-7}{8}\ge 13⋅8n−7​≥1 3n−21≥83n-21\ge 83n−21≥8 3n≥293n\ge 293n≥29 n≥293.n\ge \frac{29}{3}.n≥329​.

So

n≥10.n\ge 10.n≥10.

Compare T10T_{10}T10​ and T9T_9T9​

T10T9=3⋅n−9+19=3⋅n−89.\frac{T_{10}}{T_9}=3\cdot \frac{n-9+1}{9}=3\cdot \frac{n-8}{9}.T9​T10​​=3⋅9n−9+1​=3⋅9n−8​.

For T9≥T10T_9\ge T_{10}T9​≥T10​,

T10T9≤1\frac{T_{10}}{T_9}\le 1T9​T10​​≤1 3⋅n−89≤13\cdot \frac{n-8}{9}\le 13⋅9n−8​≤1 n−83≤1\frac{n-8}{3}\le 13n−8​≤1 n−8≤3n-8\le 3n−8≤3 n≤11.n\le 11.n≤11.

Thus the 9th term is greatest when

10≤n≤11.10\le n\le 11.10≤n≤11.

Hence the least value is

n0=10.n_0=10.n0​=10.
  1. Find the ratio of coefficient of x6x^6x6 to coefficient of x3x^3x3

Now use n=n0=10n=n_0=10n=n0​=10.

Expand

(3+6x)10.(3+6x)^{10}.(3+6x)10.

The general term is

Tr+1=(10r)310−r(6x)r.T_{r+1}=\binom{10}{r}3^{10-r}(6x)^r.Tr+1​=(r10​)310−r(6x)r.

So the coefficient of xrx^rxr is

(10r)310−r6r.\binom{10}{r}3^{10-r}6^r.(r10​)310−r6r.

Coefficient of x6x^6x6

C6=(106)3466.C_6=\binom{10}{6}3^{4}6^{6}.C6​=(610​)3466.

Coefficient of x3x^3x3

C3=(103)3763.C_3=\binom{10}{3}3^{7}6^{3}.C3​=(310​)3763.

Therefore

k=C6C3=(106)3466(103)3763.k=\frac{C_6}{C_3} =\frac{\binom{10}{6}3^4 6^6}{\binom{10}{3}3^7 6^3}.k=C3​C6​​=(310​)3763(610​)3466​.

Since

(106)=(104)=210,(103)=120,\binom{10}{6}=\binom{10}{4}=210, \qquad \binom{10}{3}=120,(610​)=(410​)=210,(310​)=120,

we get

k=210120⋅6333=74⋅(63)3=74⋅23=74⋅8=14.k=\frac{210}{120}\cdot \frac{6^3}{3^3} =\frac{7}{4}\cdot \left(\frac{6}{3}\right)^3 =\frac{7}{4}\cdot 2^3 =\frac{7}{4}\cdot 8 =14.k=120210​⋅3363​=47​⋅(36​)3=47​⋅23=47​⋅8=14.
  1. Compute k+n0k+n_0k+n0​
k+n0=14+10=24.k+n_0=14+10=24.k+n0​=14+10=24.
  1. Comparison with stored answer

Derived answer:

242424

Stored correct answer:

242424

They match.

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