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Binomial Theorem question

2022 · 25 Jun · Shift 2 · Q40
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  5. /2022 · 25 Jun · Shift 2 · Q40

Binomial Theorem question

2022 · 25 Jun · Shift 2 · Q40

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
If the sum of the co-efficient of all the positive even powers of x in the binomial expansion of (2x3+3x)10{\left( {2{x^3} + {3 \over x}} \right)^{10}}(2x3+x3​)10 is 510−β . 39{5^{10}} - \beta \,.\,{3^9}510−β.39, then β\betaβ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 83

  1. Write the general term

For

(2x3+3/x)10,(2x^3 + 3/x)^{10},(2x3+3/x)10,

the general term is

Tr+1=(10r)(2x3)10−r(3x)r.T_{r+1} = \binom{10}{r}(2x^3)^{10-r}\left(\frac{3}{x}\right)^r.Tr+1​=(r10​)(2x3)10−r(x3​)r.

Simplify:

Tr+1=(10r)210−r3rx3(10−r)−r=(10r)210−r3rx30−4r.T_{r+1} = \binom{10}{r}2^{10-r}3^r x^{3(10-r)-r} = \binom{10}{r}2^{10-r}3^r x^{30-4r}.Tr+1​=(r10​)210−r3rx3(10−r)−r=(r10​)210−r3rx30−4r.

So the power of xxx in the rrr-th choice is

30−4r.30-4r.30−4r.
  1. Find which powers are positive even powers

We need powers of the form xnx^nxn where nnn is positive and even.

Since

30−4r=2(15−2r),30-4r = 2(15-2r),30−4r=2(15−2r),

it is always even. So we only need it to be positive:

30−4r>0⇒r<7.5.30-4r > 0 \Rightarrow r < 7.5.30−4r>0⇒r<7.5.

Thus

r=0,1,2,3,4,5,6,7.r = 0,1,2,3,4,5,6,7.r=0,1,2,3,4,5,6,7.

So the required sum of coefficients is

S=∑r=07(10r)210−r3r.S = \sum_{r=0}^{7} \binom{10}{r}2^{10-r}3^r.S=r=0∑7​(r10​)210−r3r.
  1. Use the full binomial sum

We know

∑r=010(10r)210−r3r=(2+3)10=510.\sum_{r=0}^{10} \binom{10}{r}2^{10-r}3^r = (2+3)^{10} = 5^{10}.r=0∑10​(r10​)210−r3r=(2+3)10=510.

Hence

S=510−∑r=810(10r)210−r3r.S = 5^{10} - \sum_{r=8}^{10} \binom{10}{r}2^{10-r}3^r.S=510−r=8∑10​(r10​)210−r3r.
  1. Compute the omitted terms

For r=8r=8r=8:

(108)2238=45⋅4⋅38=180⋅38=60⋅39.\binom{10}{8}2^2 3^8 = 45\cdot 4\cdot 3^8 = 180\cdot 3^8 = 60\cdot 3^9.(810​)2238=45⋅4⋅38=180⋅38=60⋅39.

For r=9r=9r=9:

(109)2139=10⋅2⋅39=20⋅39.\binom{10}{9}2^1 3^9 = 10\cdot 2\cdot 3^9 = 20\cdot 3^9.(910​)2139=10⋅2⋅39=20⋅39.

For r=10r=10r=10:

(1010)20310=310=3⋅39.\binom{10}{10}2^0 3^{10} = 3^{10} = 3\cdot 3^9.(1010​)20310=310=3⋅39.

Therefore,

∑r=810(10r)210−r3r=(60+20+3)39=83⋅39.\sum_{r=8}^{10} \binom{10}{r}2^{10-r}3^r = (60+20+3)3^9 = 83\cdot 3^9.r=8∑10​(r10​)210−r3r=(60+20+3)39=83⋅39.

So

S=510−83⋅39.S = 5^{10} - 83\cdot 3^9.S=510−83⋅39.

Comparing with

510−β 39,5^{10} - \beta\,3^9,510−β39,

we get

β=83.\beta = 83.β=83.
  1. Comparison with stored answer

Stored correct answer: 838383

Our derived answer: 838383

They match.

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