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Model Paper 3 — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 33 Marks
Question
Using binomial theorem, prove that $\left(2^{3 n }-7 n -1\right)$ is divisible by $49 ,$ where $n \in N$
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Answer

To prove: $\left(2^{3 n}-7 n-1\right)$ is divisible by $49 ,$ where $n \in N$
$=\left(2^{3 n}-7 n-1\right)$
$=\left(2^3\right) n-7 n-1$
$=8^n-7 n-1$
$=(1+7)^n-7 n-1$
Now using binomial theorem..
$\Rightarrow{ }^n C_0 1^n+{ }^n C_1 1^{n-1} 7+{ }^n C_2 1^{n-2} 7^2+\ldots \ldots+{ }^n C_{n-1} 7^{n-1}+{ }^n C_n 7^n-7 n-1$
$={ }^n C_0+{ }^n C_1 7+{ }^n C_2 7^2+\ldots \ldots+{ }^n C_{n-1} 7^{n-1}+{ }^n C_n 7^n-7 n-1$
$=1+7 n+7^2\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right]-7 n-1$
$=7^2\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right]$
$=49\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right]$
$=49 K, $
$\text { where } K=\left({ }^n C_2+{ }^n C_3 7+\ldots++{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right)$
Now, $\left(2^{3 n}-7 n-1\right)=49 K$
Therefore $\left(2^{3 n}-7 n-1\right)$ is divisible by $49 .$

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