n^3 + 5n is divisible by which of the following? - Vidyadip

MCQ

$n^3 + 5n$ is divisible by which of the following?

✓
3
B
5
C
7
D
11

✓

Answer

Correct option: A.

3

3

Solution:
$P(n)=n^3+5 n$
$P(1)=1+5$
$P(1)=6$
We assume the $P(k)$ is true and divisible by 6 .
$P(k)=k^3+5 k$ is divisible by 6 and can be written as 6 c or $3 \times 2 c$
We need to prove that $P(k+1)$ is divisible by 6
$P(k+1)=(k+1)^3+5(k+1)$
$P(k+1)=k^3+1+3 k^2+3 k+5 k+5$
$P(k+1)=\left(k^3+5 k\right)+3 k^2+3 k+6$
$P(k+1)=6 c+3\left(k^2+k+2\right)$
$P(k+1)=(3 \times 2 c)+3\left(k^2+k+2\right)$
Therefore, $P(k+1)$ is definitely divisible by 3

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

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2 x)^{ n }$ be in the ratio $2: 5: 8$. Then the coefficient of the term, which is in the middle of these three terms, is $...........$.

If i < m < n, then the median of the list i, m, n is ____.

If $(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17}+\ldots \ldots$. $+20(21)^{19}= k (20)^{19}$, then $k$ is equal to

$\cos A + \sin (270^\circ + A) - \sin (270^\circ - A) + \cos (180^\circ + A) = $

Solve $\text{x}^2 + 1 = 0$.

Choose the correct answer. A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is:

If the $6^{th}$ term in the expansion of the binomial ${\left[ {\sqrt {{2^{\log (10 - {3^x})}}} + \sqrt[5]{{{2^{(x - 2)\log 3}}}}} \right]^m}$ is equal to $21$ and it is known that the binomial coefficients of the $2^{nd}$, $3^{rd}$ and $4^{th}$ terms in the expansion represent respectively the first, third and fifth terms of an $A.P$. (the symbol log stands for logarithm to the base $10$), then $x = $

$3$ boys $B_i,i = 1 , 2, 3$ and $6$ girls $G_i, i = 1 , 2, . . . , 6$ are to be seated in a row . Number of ways they can be seated so that $B_1$, $B_2$ are separated and $G_1$ , $G_2$ are also separatetd equal to

If the vertices of a triangle are $A(1,4),\,B(3,0)$ and $C(2,1),$ then the length of the median passing through $C$ is

The solution set of $8x \equiv 6(\bmod 14),\,x \in Z$, are