Principle of Mathematical Induction Test

Result

Principle of Mathematical Induction Test - 11

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$P(n)$$ is statement such that $$P(3)$$ is true. Assuming P(k) is true $$\Rightarrow$$ $$P(k+1)$$ is true for all $$k$$ $$\geq$$ $$2$$, then $$ P(n)$$ is true.

A
For all $$n$$

B
For $$n$$ $$\geq$$ $$3$$

C
For $$n$$ $$\geq$$ $$4$$

D
None of these

Solution

Given $$P (3)$$ is true.
Assume $$P(k)$$ is true $$\Rightarrow$$ $$P(k+1)$$ is true means if $$P(3)$$ is true $$\Rightarrow$$ $$P(4)$$ is true $$\Rightarrow$$ $$P(5)$$ is true and so on. So statement is true for all $$n\geq3$$.
Question 2
1 / -0

$$\displaystyle \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ......... + \frac{1}{2^n} = 1 - \frac{1}{2^n}$$ is true for

A
Only natural number n $$\geq$$ 3

B
Only natural number n $$\geq$$ 5

C
Only natural number n < 10

D
All natural number n

Solution

Let P(n)$$: \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .......... + \frac{1}{2^n} = 1 - \frac{1}{2^n}$$
Putting $$n=1, LHS = \frac{1}{2} , RHS = 1 - \frac{1}{2} = \frac{1}{2}$$ i.e., LHS $$=$$ RHS $$= \frac{1}{2}$$
$$\therefore$$ P(n) is true for n$$=$$ 1.
Suppose P(n) is true for n$$=$$ k
$$\therefore \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ......... + \frac{1}{2^k} = 1 - \frac{1}{2^k}$$
last term $$= \displaystyle \frac{1}{2^k}$$; Replacing k by k+1, last term $$=\displaystyle \frac{1}{2^{k+1}}$$
Adding $$\displaystyle \frac{1}{2^{k+1}}$$ to both sides,
$$LHS = \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ........ + \frac{1}{2^k}+ \frac{1}{2^{k+1}}$$
$$RHS = 1 - \displaystyle \frac{1}{2^k} + \frac{1}{2^{k+1}}=1-\dfrac{1}{2^k} \left ( 1 - \frac{1}{2} \right ) = 1 - \frac{1}{2^k} \cdot \frac{1}{2} = 1 - \frac{1}{2^{k+1}}$$
This shows P(n) is true for n $$=$$ k+1
Thus P(k+1) is true whenever P(k) is true
Hence, P(n) is true for all n $$\in$$ N
Question 3
1 / -0

$$1.3 + 3.5 + 5.7 + ........... + (2n -1) (2n + 1) = \displaystyle \frac{n (4n^2 + 6n -1)}{3}$$ is true for

A
Only natural number n $$\geq$$ 4

B
Only natural numbers 3 $$\leq$$ n $$\leq$$ 10

C
All natural numbers n

D
None

Solution

Le P(n) be the given statement
i.e. P(n) : 1.3 + 3.5 + 5.7 + ........... + (2n -1)(2n+1)
$$=\displaystyle \frac{n (4n^2 + 6n -1)}{3}$$
Putting $$n=1, L.H.S. = 1.3 = 3$$ and $$RHS = \displaystyle \frac{1 (4.1^2 + 6.1 -1)}{3}$$
$$ = \displaystyle \frac{4 + 6 -1}{3} = \frac{9}{3} = 3$$
LHS $$=$$ RHS
$$\therefore$$ P(n) is true for n $$=$$ 1
Assume that P(n) is true for n $$=$$ k
i.e., p(k) is true
i.e., P(k) : 1.3 + 3.5 + 5.7 + ............ + (2k-1)(2k+1)
$$= \displaystyle \frac{k (4k^2 + 6k -1)}{3}$$
Last term $$= $$ (2k -1)(2k +1)
Replacing k by (k+1), we get
$$[2(k+1)-1] [2 (k+1)+1]= (2k+1)(2k + 3)$$
$$\therefore $$ Adding (2k+1)(2k+3) on both sides.
$$\therefore LHS = 1.3 + 3.5 + 5.7 + ........... + (2k-1) (2k +1) + (2k +1) (2k +3)$$
$$R.H.S = \displaystyle \frac{k (4k^2 + 6k -1)}3{} + (2k +1) (2k +3)$$
$$ = \displaystyle \frac{(4k^3 + 6k^2 - k) + 3 (2k +1) (2k +3)}{3}$$
$$ = \displaystyle \frac{4k^3 + 18k^2 +23 +9}{3}$$
$$= \displaystyle \frac{(k+1) (4k^2 + 14 k +9)}{3}$$
$$ = \displaystyle \frac{(k+1)(4 (k+1)^2 + 6 (k+1)-1)}{3}$$ ............ (iii)
Thus, P(n) is true for n $$=$$ k + 1
$$\therefore $$ P(k+1) is true whenever P(k) is true
Hence, by P(n) is true for all n $$\in$$ N
Question 4
1 / -0

$$\forall n\in N, 2 \cdot 4^{2n + 1} + 3^{3n + 1}$$ is divisible by

A
$$7$$

B
$$5$$

C
$$11$$

D
$$209$$

Solution

For $$n = 1$$, we have
$$2 \cdot 4^{2n + 1} + 3^{3n + 1} = 2 \times 4^3 + 3^4 = 209$$, which is divisible by $$11$$
For $$n = 2$$, we have
$$2 \cdot 4^{2n + 1} + 3^{3n + 1} = 2 \times 4^5 + 3^7 = 4235$$, which is divisible by $$11$$
Hence, options (c) is true.
Question 5
1 / -0

$$\forall n\in N, 3^{3n} - 26^{n}$$ is divisible by

A
$$24$$

B
$$64$$

C
$$17$$

D
none of these

Solution

Put $$n=2$$
$$3^{3n}-26^n = 3^6-26^2 =53$$ which is not divisible by any $$24,64,17$$
Hence option 'D' is correct choice.
Question 6
1 / -0

$$7^{2n} + 3^{n - 1} \cdot 2^{3n - 3}$$ is divisible by

A
$$24$$

B
$$25$$

C
$$9$$

D
$$13$$

Solution

Let $$P(1) = 7^{2n} + 3^{n - 1} \cdot 2^{3n - 3}$$

$$P(1) = 50\Rightarrow $$ Divisible by $$25$$

Hence option 'B' is the correct choice.
Question 7
1 / -0

$$\forall n\in N; 10^{2n - 1}+1$$ is divisible by

A
$$2$$

B
$$3$$

C
$$7$$

D
$$11$$

Solution

For $$n = 1$$, we have
$$10^{2n - 1} + 1 = 10 + 1 = 11$$E, which is divisible by $$11$$.

For $$n = 2$$, we have
$$10^{2n - 1} + 1 = 10^3 + 1 = 1001$$, which is divisible by $$11$$.

$$\forall n\in N; 10^{2n - 1}+1$$ is divisible by 11.

Hence, option (d) is correct.
Question 8
1 / -0

If $$x^n - 1$$ is divisible by $$x - k$$, then the least positive integral value of $$k$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

if $$f(x)=x^n-1$$ is divisible by $$x-k$$
Then $$f(k)=0$$
Therefore, $$k^n=1$$
and thus least positive integral value of k is 1
Question 9
1 / -0

$$\forall n\in N, n^4$$ is less than

A
$$10^n$$

B
$$4n$$

C
$$4^n$$

D
$$10^{10}$$

Solution

Let $$y =n^4$$
put $$n=10\Rightarrow y = 10^4=10000>4\times 10, 4^4$$
Hence option B and C discarded.
Now put $$n = 10^3\Rightarrow y = 10^{12}>10^{10}$$. thus option D is also incorrect.
Hence option 'A' is correct.
Question 10
1 / -0

If $$n$$ $$ \in$$ N, then $$x^{2n - 1} + y^{2n - 1}$$ is divisible by

A
$$x + y$$

B
$$x - y$$

C
$$x^2 + y^2$$

D
none of these

Solution

Le4 the given statement be $$P\left( n \right) $$
$$P\left( n \right) ={ x }^{ 2n-1 }+{ y }^{ 2n-1 }$$
We check $$P\left( n \right) $$ for $$n=1$$
$$P\left( 1 \right) ={ x }^{ 2-1 }+{ y }^{ 2-1 }=x+y$$
Thus $$P\left( 1 \right) $$ is divisible by $$x+y$$

Let us assume that $$P\left( n \right) $$ is divisible by $$x+y$$ when $$n=k$$
Now for $$n=k+1$$ we have
$$P\left( k+1 \right) ={ x }^{ 2k+1 }+{ y }^{ 2k+1 }$$
$$\Rightarrow P\left( k+1 \right) =\left( { x }^{ 2 } \right) \left( { x }^{ 2k-1 } \right) +\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 } \right) $$
$$\Rightarrow P\left( k+1 \right) =\left( { x }^{ 2 } \right) \left( { x }^{ 2k-1 }-{ y }^{ 2 } \right) \left( { x }^{ 2k-1 }+{ y }^{ 2 } \right) \left( { x }^{ 2k-1 } \right) +\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 } \right) $$
$$\Rightarrow P\left( k+1 \right) =\left( { x }^{ 2 }-{ y }^{ 2 } \right) \left( { x }^{ 2k-1 } \right) +\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 }+{ x }^{ 2k-1 } \right) $$
$$\Rightarrow P\left( k+1 \right) =\left( { x }-{ y } \right) \left( { x }+{ y } \right) \left( { x }^{ 2k-1 } \right) +\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 }+{ x }^{ 2k-1 } \right) $$

Now $$\left( { x }-{ y } \right) \left( { x }+{ y } \right) \left( { x }^{ 2k-1 } \right) $$ and $$\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 }+{ x }^{ 2k-1 } \right) $$ are divisible by $$x+y$$
so $${ x }^{ 2k+1 }+{ y }^{ 2k+1 }$$ is also divisible by $$x+y$$
So $${ x }^{ 2n-1 }+{ y }^{ 2n-1 }$$ is divisible by $$x+y$$ when $$n=k+1$$ if it is divisible by $$x+y$$ when $$n=k$$
so by principle of mathematical induction, $${ x }^{ 2n-1 }+{ y }^{ 2n-1 }$$ is divisible by $$x+y$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Principle of Mathematical Induction Test - 11

Result

Principle of Mathematical Induction Test - 11

Score

-

Rank

-

SHARING IS CARING

Question 1

For all $$n$$

For $$n$$ $$\geq$$ $$3$$

For $$n$$ $$\geq$$ $$4$$

None of these

Solution

Question 2

Only natural number n $$\geq$$ 3

Only natural number n $$\geq$$ 5

Only natural number n < 10

All natural number n

Solution

Question 3

Only natural number n $$\geq$$ 4

Only natural numbers 3 $$\leq$$ n $$\leq$$ 10

All natural numbers n

None

Solution

Question 4

$$7$$

$$5$$

$$11$$

$$209$$

Solution

Question 5

$$24$$

$$64$$

$$17$$

none of these

Solution

Question 6

$$24$$

$$25$$

$$9$$

$$13$$

Solution

Question 7

$$2$$

$$3$$

$$7$$

$$11$$

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

$$10^n$$

$$4n$$

$$4^n$$

$$10^{10}$$

Solution

Question 10

$$x + y$$

$$x - y$$

$$x^2 + y^2$$

none of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.