Let P(n)=2^ 3 n -7 n-1 then P(n) is divisible by: - Vidyadip

MCQ

Let $P(n)=2^{3 n}-7 n-1$ then $P(n)$ is divisible by:

A
63
B
36
✓
49
D
25

✓

Answer

Correct option: C.

49

49

Solution:
$P(n)=2^{3 n}-7 n-1=-1-7 n+(1+7)^n$
$\Rightarrow P(n)=-1-7 n+\left(1+n C_1 7+n C_2 7^2+\ldots+n C_n 7^n\right)=n C_2 7^2+\ldots+n C_n 7^n$
$\Rightarrow P(n)=7^2\left(n C_2+n C_3 7+\ldots+n C_n 7^{n-2}\right)$
Therefore, $P(n)$ is divisible by 49 .

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from PRINCIPLE OF MATHEMATICAL INDUCTION→PRINCIPLE OF MATHEMATICAL INDUCTION — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $

If $\frac{1}{x} + x = 2\cos \theta ,$ then ${x^n} + \frac{1}{{{x^n}}}$ is equal to

Choose the correct answer. If e is the eccentricity of the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b}),$ then:

The value of the limit of $\frac{{{x^3} - {x^2} - 18}}{{x - 3}}$ as $x$ tends to $3$ is

If $|{z_1} + {z_2}| = |{z_1} - {z_2}|$, then the difference in the amplitudes of ${z_1}$ and ${z_2}$ is

An integer is chosen at random and squared. The probability that the last digit of the square is $1$ or $5$ is

Choose the correct answer.If A and B are coefficient of $x^n$ in the expansions of $(1 + x)^{2n}$ and $(1 + x)^{2n – 1}$ respectively, then $\frac{\text{A}}{\text{B}}$ equals:
Hint: $\frac{\text{A}}{\text{B}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=2$

If the distance between the foci of an ellipse be equal to its minor axis, then its eccentricity is

The sum of all the coefficients in the binomial expansion of ${({x^2} + x - 3)^{319}}$ is

The sum of the series $\sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !}$ is equal to :