Let a-41-401-1 and for each n-2- let bn- nC1- nC2- a- nC3- a2- nCn- an-1- Then the value of b2020-b2019 is

B_n= ^nC_1+ ^nC_2· a+ ^nC_3· a^2+ ^nC_n· a^n 1 =1a[ ^nC_1· a+ ^nC_2· a^2+ ^nC_3· a^3+ ^nC_n· a^n] =1a[ ^nC_0+ ^nC_1· a+ ^nC_2· a^2+ ^nC_3· a^3+ ^nC_n· a^n 1] =1 ...

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Byju's Answer

Standard XIII

Mathematics

Binomial Expression

Let a=41/401-...

Question

Let $a = (4^{1 / 401} - 1)$ and for each $n \geq 2,$ let $b_{n} =^{n} C_{1} +^{n} C_{2} \cdot a +^{n} C_{3} \cdot a^{2} +^{n} C_{n} \cdot a^{n - 1} .$ Then the value of $b_{2020} - b_{2019}$ is

4^{2020 / 401}

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4^{2019 / 401}

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- 4^{2020 / 401}

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- 4^{2019 / 401}

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Solution

The correct option is B $4^{2019 / 401}$
$b_{n} =^{n} C_{1} +^{n} C_{2} \cdot a +^{n} C_{3} \cdot a^{2} +^{n} C_{n} \cdot a^{n - 1}$
$= \frac{1}{a} [^{n} C_{1} \cdot a +^{n} C_{2} \cdot a^{2} +^{n} C_{3} \cdot a^{3} +^{n} C_{n} \cdot a^{n}]$
$= \frac{1}{a} [^{n} C_{0} +^{n} C_{1} \cdot a +^{n} C_{2} \cdot a^{2} +^{n} C_{3} \cdot a^{3} +^{n} C_{n} \cdot a^{n} - 1]$
$= \frac{1}{a} ((1 + a)^{n} - 1)$
Now
$b_{2020} - b_{2019} = \frac{1}{a} ((1 + a)^{2020} - 1) - \frac{1}{a} ((1 + a)^{2019} - 1)$
$= \frac{1}{a} [(1 + a)^{2020} - (1 + a)^{2019}]$
$= \frac{1}{a} [(1 + a)^{2019} (1 + a - 1)]$
$= (1 + a)^{2019}$
$= (1 + 4^{1 / 401} - 1)^{2019}$
$= 4^{2019 / 401}$

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Let a=(41/401−1) and for each n≥2, let bn= nC1+ nC2⋅a+ nC3⋅a2+ nCn⋅an−1. Then the value of b2020−b2019 is

Let $a = (4^{1 / 401} - 1)$ and for each $n \geq 2,$ let $b_{n} =^{n} C_{1} +^{n} C_{2} \cdot a +^{n} C_{3} \cdot a^{2} +^{n} C_{n} \cdot a^{n - 1} .$ Then the value of $b_{2020} - b_{2019}$ is