The value of 20C0- 20C1- 20C2- 20C3- 20C10 is

We know that, nbsp; (1+x)^20 nbsp;= nbsp; ^20C_0+ ^20C_1x+ ^20C_2x^2+ ^20C_3x^3+…+ ^20C_20x^20 Putting x= 1, we get ^20C_0 ^20C_1+ ^20C_2 ^20C_3+…+ ^2 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

The value of ...

Question

The value of $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10}$ is

\frac{^{20} C_{10}}{2}

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- \frac{^{20} C_{10}}{2}

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^{20} C_{10}

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-^{20} C_{10}

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Solution

The correct option is A $\frac{^{20} C_{10}}{2}$
We know that,
$(1 + x)^{20} =^{20} C_{0} +^{20} C_{1} x +^{20} C_{2} x^{2} +^{20} C_{3} x^{3} + \dots +^{20} C_{20} x^{20}$
Putting $x = - 1$ , we get
$^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{20} = 0 \Rightarrow 2 (^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots) +^{20} C_{10} = 0 (∵^{20} C_{r} =^{20} C_{20 - r}) \Rightarrow (^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots) = - \frac{^{20} C_{10}}{2} ∴^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10} = \frac{^{20} C_{10}}{2}$

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The value of 20C0− 20C1+ 20C2− 20C3+…+ 20C10 is

The value of $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10}$ is