Let a - 41-401 -1 and for each n - 2- let bn - nC1 - nC2- a-nC3- a2 - -nCn- an-1- If the value of b2006 - b2005 is 4k where k - then the value of k is

A = 4^1/401 1 ⇒ 1 + a = 4^1/401 b_n = ^nC_1 + ^nC_2· a+ ^nC_3· a^2 +… +^n_C_n· a^n 1 = 1a[ ^nC_1· a + ^nC_2· a^2 + ^nC_3· a^3 + nbsp; … + ^nC_n· a^n ] = 1a ...

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Byju's Answer

Standard XII

Mathematics

Binomial Expression

Let a = 41/40...

Question

Let $a = (4^{1 / 401} - 1)$ and for each $n \geq 2,$ let $b_{n} =^{n} C_{1} +^{n} C_{2} \cdot a +^{n} C_{3} \cdot a^{2} + \dots +^{n} C_{n} \cdot a^{n - 1} .$ If the value of $(b_{2006} - b_{2005})$ is $4^{k}$ where $k \in N,$ then the value of $k$ is

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Solution

$a = 4^{\frac{1}{401}} - 1 \Rightarrow 1 + a = 4^{\frac{1}{401}}$

$b_{n} =^{n} C_{1} +^{n} C_{2} \cdot a +^{n} C_{3} \cdot a^{2} + \dots + {^{n}}_{C_{n}} \cdot a^{n - 1}$
$= \frac{1}{a} [^{n} C_{1} \cdot a +^{n} C_{2} \cdot a^{2} +^{n} C_{3} \cdot a^{3} + \dots +^{n} C_{n} \cdot a^{n}]$
$= \frac{1}{a} [{(1 + a)}^{n} - 1]$
$\Rightarrow b_{n} = \frac{1}{a} [4^{n / 401} - 1] \dots (1)$

$b_{2006} = \frac{1}{a} [{(4^{1 / 401})}^{2006} - 1]; b_{2005} = \frac{1}{a} [{(4^{1 / 401})}^{2005} - 1]$

$(b_{2006} - b_{2005}) = \frac{1}{a} [{(4^{1 / 401})}^{2006} - {(4^{1 / 401})}^{2005}] = \frac{1}{a} {(4^{1 / 401})}^{2005} [4^{1 / 401} - 1]      = a$
$= (4^{2005 / 401}) = 4^{5} \equiv 4^{k}$
$∴ k = 5$

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Let a=(41/401−1) and for each n≥2, let bn=nC1+nC2⋅a+nC3⋅a2+⋯+nCn⋅an−1. If the value of (b2006−b2005) is 4k where k∈N, then the value of k is

Let $a = (4^{1 / 401} - 1)$ and for each $n \geq 2,$ let $b_{n} =^{n} C_{1} +^{n} C_{2} \cdot a +^{n} C_{3} \cdot a^{2} + \dots +^{n} C_{n} \cdot a^{n - 1} .$ If the value of $(b_{2006} - b_{2005})$ is $4^{k}$ where $k \in N,$ then the value of $k$ is