Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 10

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Question 1
1 / -0

If $$n(n^{2}-1)$$ is divisible by $$24$$, then which of the following statements is true?

A
$$n$$ can be any odd integral value.

B
$$n$$ can be any integral value.

C
$$n$$ can be any even integral value.

D
$$n$$ can be any rational number.

Solution

$$n({ n }^{ 2 }-1)\quad =\quad n*(n-1)*(n+1)$$
If $$n$$ is even, then $$n-1$$ and $$n+1$$ will be odd, therefore $$n({ n }^{ 2 }-1)$$ is not divisible by 4 and therefore not divisible by 24.
Hence $$n$$ has to be an odd integer.
Question 2
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If $$\forall m\in N$$, then $$11^{m+ 2}+12^{2m-1}$$ is divisible by

A
$$121$$

B
$$132$$

C
$$133$$

D
None of these

Solution

To find the divisor of $${ 11 }^{ m+2 }+{ 12 }^{ 2m-1 }$$ by mathematic induction, the first step is to check for the smallest natural number, i.e; for $$m=1$$. So, this reduces to $${ 11 }^{ 3 }+{ 12 }^{ 1 }$$ or $${ 11 }^{ 4 }+1$$.
So, the number when divided by $$11$$ leaves remainder 1.
So, we can knock out options $$A$$ and $$B$$ as $$121$$ as well as $$132$$ are both divisible by 11 and hence their multiples will always be divisible by 11.
Now, we have to check the divisibility of $${ 11 }^{ m+2 }+{ 12 }^{ 2m-1 }$$ by $$133$$. For $$m=1$$, $${ 11 }^{ 4 }+1$$ is not divisible by $$133$$.
So, we can knock out option $$C$$.
Hence, $$D$$ is correct.
Question 3
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If $$n$$ is an even number, then the digit in the units place of $$2^{2n}+1$$ will be

A
$$5$$

B
$$7$$

C
$$6$$

D
$$1$$

Solution

Since $${ 2 }^{ 2n }$$ is even therefore $${ 2 }^{ 2n }+1$$ is odd,
therefore digit at unit place should be odd, rejecting option 3.
Put n = 2, we get $${ 2 }^{ 2n } +1$$ = 17,
hence digit should be 7
Question 4
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Let $$P(n):1+\displaystyle \frac{1}{4}+\frac{1}{9}+\ldots..+\frac{1}{n^{2}}<2-\frac{1}{n}$$ is true for

A
$$\forall n\in N$$

B
$$n=1$$

C
$${n>1,\forall n\in N}$$

D
$$n>2$$

Solution

For $$n=2,P\left( n \right) =1+\dfrac { 1 }{ { 2 }^{ 2 } } =1+\dfrac { 1 }{ 4 } =\dfrac { 5 }{ 4 } =1.25<2-\dfrac { 1 }{ 2 } =\dfrac { 3 }{ 2 } =1.5\\ P(3)=1+\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 9 } =1.36<1.67$$
$$\therefore P(2), P(3)$$ is true and so on.
Let us assume $$P(n)$$ is true.
$$P\left( n+1 \right) =1+\dfrac { 1 }{ 4 } +.....+\dfrac { 1 }{ { n }^{ 2 } } +\dfrac { 1 }{ { \left( n+1 \right) }^{ 2 } } <2-\dfrac { 1 }{ n } +\dfrac { 1 }{ { \left( n+1 \right) }^{ 2 } } \\ =2-\left\{ \dfrac { 1 }{ n } -\dfrac { 1 }{ { \left( n+1 \right) }^{ 2 } } \right\} =2-\dfrac { { n }^{ 2 }+n+1 }{ n{ \left( n+1 \right) }^{ 2 } } =2-\dfrac { { n }^{ 2 }+n }{ n{ \left( n+1 \right) }^{ 2 } } -\dfrac { 1 }{ n{ \left( n+1 \right) }^{ 2 } } $$
$$=2-\dfrac { 1 }{ n+1 } -\dfrac { 1 }{ n{ \left( n+1 \right) }^{ 2 } } <2-\dfrac { 1 }{ n+1 } $$
$$\therefore P(n+1)$$ is true.
Thus, $$P(n)$$ is true for all $$n>1.$$
Question 5
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Let $$x > -1$$, then statement $$p(n):(1 + x)^{n} > 1 + nx$$, where $$ n \in N$$ is true for

A
For all $$n \epsilon N$$.

B
For all $$n > 1$$.

C
For all $$n > 1$$, provided $$x \neq 0$$.

D
For all $$n > 2$$.

Solution

p(1) :$$ (1 + x)^{1}$$ > 1 + x is false
p(2) :$$ (1 + x)^{2} > 1 + 2x \Rightarrow x^{2}$$ > 0 is true when $$x \neq 0$$
p(3) :$$ (1 + x)^{3} > 1 + 3x \Rightarrow x^{2}$$ > 0 is true when$$ x \neq 0$$
Let p(k) $$ (1 + x)^{k}$$ > 1 + k x is true for some $$k \epsilon N$$, k >1
$$\Rightarrow (1+x)^{k+1}$$ >(1+kx) (1+x)
$$\Rightarrow (1+x)^{k+1} > 1+(k+1)x+kx^{2} > 1 +(k+1)x, x \neq 0$$
$$\Rightarrow$$p(k+1) is true whenever p(k) is true
so by principle of mathematical induction p(n) is true for all n > 1 provided $$x \neq 0$$
Question 6
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$$n(n+1) (n+5)$$ is a multiple of $$3$$ is true for

A
All natural numbers $$n > 5$$

B
Only natural number $$3$$ $$\leq$$ $$n < 15$$

C
All natural numbers $$n$$

D
None

Solution

Let the statement be denoted by p(n) i.e.,
P(n) : n(n+1) (n+5) is a multiple of 3
For n $$=$$ 1, n(n+1) (n+5) $$=$$ 1.2.6 $$=$$ 12 $$=$$ 3.4
P(n) is true for n $$=$$ 1
Suppose p(k) is true for n $$=$$ k i.e.
k(k+1) (k+5) =3m (let) or k$$^3$$ + 6k$$^2$$ + 5k $$=$$ 3m ........... (i)
Replacing k by k+1, we get
(k+1) (k+2) (k+6) $$=$$ k (k$$^2$$ +8k +12) + (k$$^2$$ + 8k + 12)
$$k^3 + 9k^2 +20 k +12 = (k^3 + 6k^2 + 5k) + (3k^2 +15k + 12)$$
$$= 3m + 3k^2 +15k + 12$$ [from (i)]
$$= 3(m + k^2 + 5k +4)$$
i.e. (k+1) (k+2) (k+6) is a multiple of 3
i.e. P(k+1) is multiple of 3, if P(k) is a multiple of 3
i.e. P(k+1) is true whenever P(k) is true.
Hence P(n) is true for all n $$\in$$ N
Question 7
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$$1.2 +2.2^2 +3.2^3+ .................+ n.2^n = (n-1)2^{n+1} + 2$$ is true for

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

B
All natural number $$n$$

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

D
None

Solution

Let P(n) be the given statement i.e. $$P(n) : 1.2 + 2.2^2 + 3.2^3+ ......... + n.2^n = (n-1)2^{n+1}+2$$
Putting n $$=$$ 1, LHS $$=$$ 1.2 $$=$$ 2; RHS $$=$$ 0+2 $$=$$ 2
$$\therefore$$ P(n) is true for $$n=1$$
Assume that P(n) is true for $$n=k$$
i.e. P(k) is true i.e. $$1.2^2+ 2.2^2 + 3.2^3 + ........... + k.2$$ $$^k$$ Replacing k by k + 1, we get the next term $$=(k+1)2^{k+1}$$
Adding it to both sides
$$LHS = 1.2 +2.2^2 + 3.2^3+ ......... + k.2^{k}+ (k+1)2^{k+1}$$
$$RHS = (k-1)2^{k+1}+2 + (k+1)2^{k+1}$$
$$=2^{k+1}[k-1+k+1]+2=2k2^{k+1}+2=(k+1-1)2^{k+1+1}+2$$
This proves P(n) 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 8
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If $$n\in N$$, then $$n(n^2-1)$$ is divisible by

A
$$6$$

B
$$16$$

C
$$26$$

D
$$24$$

Solution

$${ n }{ (n }^{ 2 }-1)={ n }{ (n }-1){ (n }+1)$$

One of the $$n$$, $$n+1$$ and $$n-1$$ will be a multiple of $$3$$.

Since $$n-1$$, $$n$$ and $$n+1$$ are three consecutive integers, therefore at least one of them will be divisible by $$2$$.

Therefore $${ n }{ (n }^{ 2 }-1)$$ is divisible by $$6$$.

Option d can be rejected by putting $$n = 2$$
Question 9
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Using mathematical induction,
$$\displaystyle \left ( 1 - \frac{1}{2^2} \right ) \left ( 1 - \frac{1}{3^2} \right ) \left ( 1 - \frac{1}{4^2} \right ) ......... \left ( 1 - \frac{1}{(n + 1)^2} \right )$$

A
$$\displaystyle \frac{n +2}{2(n + 1)}$$

B
$$\displaystyle \frac{n - 2}{2 (n - 1)}$$

C
$$\displaystyle \frac{n + 3}{2 (n + 1)}$$

D
$$\displaystyle \frac{n}{2 (n + 1)}$$

Solution

$$\displaystyle \left ( 1 - \dfrac{1}{2^2} \right ) \left ( 1 - \dfrac{1}{3^2} \right ) \left ( 1 - \dfrac{1}{4^2} \right ) ......... \left ( 1 - \dfrac{1}{(n + 1)^2} \right )$$
nth term is $$1-\dfrac{1}{(n+1)^2}$$
for n=1 =>$$\dfrac{1}{(n+1)^2}$$=$$1-\dfrac{1}{4}$$ =$$\dfrac{3}{4}$$
for n-1 series will be
$$\left( 1-\dfrac { 1 }{ 2^{ 2 } } \right) \left( 1-\dfrac { 1 }{ 3^{ 2 } } \right) \left( 1-\dfrac { 1 }{ 4^{ 2 } } \right) .........\left( 1-\dfrac { 1 }{ (n+1-1)^{ 2 } } \right) \left( 1-\dfrac { 1 }{ (n+1)^{ 2 } } \right) $$
On simplyfying
$$\left( \dfrac { 3 }{ { 2 }^{ 2 } } \right) \left( \dfrac { 8 }{ { 3 }^{ 2 } } \right) \left( \dfrac { 15 }{ { 4 }^{ 2 } } \right) ........\left( \dfrac { (n+1)(n-1) }{ { n }^{ 2 } } \right) \left( \dfrac { n(n+2) }{ { (n+1) }^{ 2 } } \right) $$
OR
$$\left( \dfrac { 3 }{ { 2 }^{ 2 } } \right) \left( \dfrac { 2*4 }{ { 3 }^{ 2 } } \right) \left( \dfrac { 3*5 }{ { 4 }^{ 2 } } \right) ........\left( \dfrac { (n+1)(n-1) }{ { n }^{ 2 } } \right) \left( \dfrac { n(n+2) }{ { (n+1) }^{ 2 } } \right) $$
Now all terms will cancel out with each other from denominator to numinator only $$\dfrac{1}{2}$$ from first term and $$(\dfrac{n+2}{n+1})$$ from last term will remain
Therefore, Answer is $$\dfrac{1}{2}(\dfrac{n+2}{n+1})$$
Question 10
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For all positive integers $$n$$, $$P(n)$$ is true , and $$2^{n-2}>3n$$, then which of the following is true?

A
$$P(3)$$ is true.

B
$$P(5)$$ is true.

C
If $$P(m)$$ is true then $$P(m + 1)$$ is also true.

D
If $$P(m)$$ is true then $$P(m + 1)$$ is not true.

Solution

$$2^{n-2}>3n$$
Put $$n=3$$
$$2\ngtr 6$$
Hence, P(3) is not true.

Put $$n=5$$
$$8\ngtr 15$$
Hence, P(3) is not true.

Let $$P(m)$$ is true i.e.
$$2^{m-2}>3m$$

Now, we will check for $$P(m+1)$$
Consider , $$2^{m-1}$$
$$=2^{m-2}.2$$
$$>2(3m)=6m=3m+3m $$
$$>3m+3$$
Hence, $$2^{m-1}>3(m+1)$$
Hence, $$P(m+1)$$ is true.

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Re-Attempt Weekly Quiz Competition

Principle of Mathematical Induction Test - 10

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Principle of Mathematical Induction Test - 10

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SHARING IS CARING

Question 1

$$n$$ can be any odd integral value.

$$n$$ can be any integral value.

$$n$$ can be any even integral value.

$$n$$ can be any rational number.

Solution

Question 2

$$121$$

$$132$$

$$133$$

None of these

Solution

Question 3

$$5$$

$$7$$

$$6$$

$$1$$

Solution

Question 4

$$\forall n\in N$$

$$n=1$$

$${n>1,\forall n\in N}$$

$$n>2$$

Solution

Question 5

For all $$n \epsilon N$$.

For all $$n > 1$$.

For all $$n > 1$$, provided $$x \neq 0$$.

For all $$n > 2$$.

Solution

Question 6

All natural numbers $$n > 5$$

Only natural number $$3$$ $$\leq$$ $$n < 15$$

All natural numbers $$n$$

None

Solution

Question 7

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

All natural number $$n$$

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

None

Solution

Question 8

$$6$$

$$16$$

$$26$$

$$24$$

Solution

Question 9

$$\displaystyle \frac{n +2}{2(n + 1)}$$

$$\displaystyle \frac{n - 2}{2 (n - 1)}$$

$$\displaystyle \frac{n + 3}{2 (n + 1)}$$

$$\displaystyle \frac{n}{2 (n + 1)}$$

Solution

Question 10

$$P(3)$$ is true.

$$P(5)$$ is true.

If $$P(m)$$ is true then $$P(m + 1)$$ is also true.

If $$P(m)$$ is true then $$P(m + 1)$$ is not true.

Solution

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