Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 11

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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 $x+y$

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Principle of Mathematical Induction Test - 11

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Principle of Mathematical Induction Test - 11

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Question 1

For all nnn

For nnn ≥\geq≥ 333

For nnn ≥\geq≥ 444

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

777

555

111111

209209209

Solution

Question 5

242424

646464

171717

none of these

Solution

Question 6

242424

252525

999

131313

Solution

Question 7

222

333

777

111111

Solution

Question 8

111

222

333

444

Solution

Question 9

10n10^n10n

4n4n4n

4n4^n4n

101010^{10}1010

Solution

Question 10

x+yx + yx+y

x−yx - yx−y

x2+y2x^2 + y^2x2+y2

none of these

Solution

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For all $n$

For $n$ $\geq$ $3$

For $n$ $\geq$ $4$

Only natural number n $\geq$ 3

Only natural number n $\geq$ 5

Only natural number n $\geq$ 4

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

$7$

$5$

$11$

$209$

$24$

$64$

$17$

$24$

$25$

$9$

$13$

$2$

$3$

$7$

$11$

$1$

$2$

$3$

$4$

$10^n$

$4n$

$4^n$

$10^{10}$

$x + y$

$x - y$

$x^2 + y^2$