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Binomial Theorem question

2021 · 22 Jul · Shift 2 · Q45
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  5. /2021 · 22 Jul · Shift 2 · Q45

Binomial Theorem question

2021 · 22 Jul · Shift 2 · Q45

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
If the constant term, in binomial expansion of (2xr+1x2)10{\left( {2{x^r} + {1 \over {{x^2}}}} \right)^{10}}(2xr+x21​)10 is 180, then r is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 8

  1. Consider the general term in the expansion of (2xr+1x2)10.\left(2x^r+\frac{1}{x^2}\right)^{10}.(2xr+x21​)10.

Using the binomial theorem, the general term is Tk+1=(10k)(2xr)10−k(1x2)k,T_{k+1}=\binom{10}{k}(2x^r)^{10-k}\left(\frac{1}{x^2}\right)^k,Tk+1​=(k10​)(2xr)10−k(x21​)k, where k=0,1,2,…,10k=0,1,2,\dots,10k=0,1,2,…,10.

  1. Simplify this term: Tk+1=(10k)210−kxr(10−k)x−2kT_{k+1}=\binom{10}{k}2^{10-k}x^{r(10-k)}x^{-2k}Tk+1​=(k10​)210−kxr(10−k)x−2k =(10k)210−kx10r−k(r+2).=\binom{10}{k}2^{10-k}x^{10r-k(r+2)}.=(k10​)210−kx10r−k(r+2).

  2. For the constant term, the power of xxx must be zero. So, 10r−k(r+2)=0.10r-k(r+2)=0.10r−k(r+2)=0. This gives 10r=k(r+2).10r=k(r+2).10r=k(r+2).

  3. The coefficient of the constant term is given to be 180180180. So we need (10k)210−k=180.\binom{10}{k}2^{10-k}=180.(k10​)210−k=180.

Now test integer values of kkk from 000 to 101010:

  • k=8k=8k=8 gives (108)22=45⋅4=180.\binom{10}{8}2^{2}=45\cdot 4=180.(810​)22=45⋅4=180. So k=8k=8k=8.

  1. Substitute k=8k=8k=8 into the exponent condition: 10r=8(r+2)10r=8(r+2)10r=8(r+2) 10r=8r+1610r=8r+1610r=8r+16 2r=162r=162r=16 r=8.r=8.r=8.

  2. Therefore, the required integer is 8.\boxed{8}.8​.

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