Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 18

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Question 1
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Let $$a, b, c$$ and $$d$$ be any four real numbers. Then, $$a^n+b^n=c^n+d^n$$ holds for any natural number $$n$$, if

(This question has some ambiguity, but appeared in WBJEE 2015 exam).

A
$$a+b = c+d$$

B
$$a-b=c-d$$

C
$$a+b=c+d, a^2+b^2=c^2+d^2$$

D
$$a-b=c-d, a^2-b^2=c^2-d^2$$

Solution

Option A
$$a+b=c+d$$ ....Given
Squaring on both sides, we get
$$(a+b)^2 = (c+d)^2$$
$$a^2 +b^2 + 2ab = c^2+d^2+2cd$$
$$\therefore a^n+b^n \neq c^n+d^n$$

Option B
$$a-b=c-d$$ ...Given
Squaring on both sides, we get
$$(a-b)^2 = (c-d)^2$$
$$a^2 +b^2 - 2ab = c^2+d^2-2cd$$
$$\therefore a^n+b^n \neq c^n+d^n$$

Option C
$$a+b=c+d$$ ...(i)
$$a^2+b^2 = c^2+d^2$$ ....(ii)
Consider $$(a+b)^2 = (c+d)^2$$
$$a^2 +b^2 + 2ab = c^2+d^2+2cd$$
$$\Rightarrow 2ab = 2cd$$ ...from (ii)
$$\Rightarrow ab=cd$$
So, $$a^n+b^n \neq c^n+d^n$$ for all $$n \epsilon N$$.

Option D
$$a-b=c-d$$ ...(i)
$$a^2-b^2 = c^2-d^2$$ ....(ii)
Consider $$a^2-b^2 = c^2-d^2$$
$$\Rightarrow (a-b)(a+b) = (c-d)(c+d)$$
$$\Rightarrow a+b = c+d$$ ...from (i) ....(iii)
Adding eq(i) and (ii), we get
$$2a=2c$$
$$\Rightarrow a=c$$
$$\therefore b=d$$ ...from (iii)
So, $$a^n+b^n=c^n+d^n$$ for all $$n \epsilon N$$.
Question 2
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For any $$+$$ve integer $$n, n^3 + 2n$$ is always divisible by

A
$$3$$

B
$$7$$

C
$$5$$

D
$$6$$

Solution

Let $$P(n)=n^3+2n$$
Now $$P(1)=1^3+2(1)=3$$, divisible by $$3$$
$$P(2)=2^3+2(4)=12$$, divisible by $$3$$ and $$6$$
$$P(3)=3^3+2(3)=33$$, divisible by $$3$$

Hence $$n^3+2n$$ is divisible by $$3$$
Question 3
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The last digit in $$\displaystyle { 7 }^{ 300 }$$ is:

A
7

B
9

C
1

D
3

Solution

We go on writing the units digit in the first few terms for the expansion of $$7$$.
$$7^1 \rightarrow 7$$
$$7^2 \rightarrow 9$$
$$7^3 \rightarrow 3$$
$$7^4 \rightarrow 1$$
$$7^5 \rightarrow 7$$
And the terms will keep on repeating.
Thus, the trend repeats itself in multiples of $$4$$.
Since $$300$$ is divisible by $$4$$, the units digit of $$7^{300}$$ would be $$1.$$
Question 4
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For any integer $$n\ge 1$$, the sum $$\displaystyle\sum _{ k=1 }^{ n }{ k\left( k+2 \right) } $$ is equal to

A
$$\dfrac { n\left( n+1 \right) \left( n+2 \right) }{ 6 } $$

B
$$\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } $$

C
$$\dfrac { n\left( n+1 \right) \left( 2n+7 \right) }{ 6 } $$

D
$$\dfrac { n\left( n+1 \right) \left( 2n+9 \right) }{ 6 } $$

Solution

Now, $$\displaystyle\sum _{ k=1 }^{ n }{ k\left( k+2 \right) } $$
$$=\displaystyle\sum _{ k=1 }^{ n }{ { k }^{ 2 }+2k } =\displaystyle\sum _{ k=1 }^{ n }{ { k }^{ 2 } } +2k\displaystyle\sum _{ k=1 }^{ n }{ k } $$
$$=\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } +\dfrac { 2n\left( n+1 \right) }{ 2 } $$
$$=n\left( n+1 \right) \dfrac { 2n+1 }{ 6 } +1$$
$$=\dfrac { n\left( n+1 \right) \left( 2n+7 \right) }{ 6 } $$
Question 5
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Let $$A = \begin{pmatrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}$$. Then for positive integer $$n, A^{n}$$ is

A
$$\begin{pmatrix}1 & n & n^{2}\\ 0 & n^{2} & n\\ 0 & 0 & n\end{pmatrix}$$

B
$$\begin{pmatrix}1 & n & n\left (\dfrac {n + 1}{2}\right )\\ 0 & 1 & n\\ 0 & 0 & 1\end{pmatrix}$$

C
$$\begin{pmatrix}1 & n^{2} & n\\ 0 & n & n^{2}\\ 0 & 0 & n^{2}\end{pmatrix}$$

D
$$\begin{pmatrix}1 & n & 2n - 1\\ 0 & \dfrac {n + 1}{2} & n^{2}\\ 0 & 0 & \dfrac {n + 1}{2}\end{pmatrix}$$

Solution

$$A= \begin{pmatrix}
1& 1& 1\\
0& 1& 1\\
0& 0&1
\end{pmatrix}$$
$$A^{2}= \begin{pmatrix}
1& 2&3 \\
0& 1&2 \\
0& 0&1
\end{pmatrix}$$
$$A^{3}= \begin{pmatrix}
1& 3&6 \\
0& 1&3 \\
0& 0&1
\end{pmatrix}$$
There using mathematical induction we get that
$$A^{n}= \begin{pmatrix}
1& n& \frac{n(n+1)}{2} \\
0& 1&n \\
0& 0&1
\end{pmatrix}$$
Question 6
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The integer next above $$(\sqrt{3}+1)^{2n}$$ contains

A
$$2^{n+1}$$ as a factor

B
$$2^{n+2}$$ as a factor

C
$$2^{n+3}$$ as a factor

D
$$2^{n}$$ as a factor

Solution

We have,

$$(\sqrt{3}+1)^{2n}$$

$$=$$ $$\{(\sqrt{3}+1)^{2}\}^n$$

$$=(4+2\sqrt{3})^n$$

$$=2^n(2+\sqrt{3})^n$$.

So it is clear that the integer next above the given number contains $$2^n$$ as a factor.
Question 7
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If $${a_{1,}}{a_{2,}}{a_{3,}}........{a_{2n + 1}}$$ are in A.P. then$$\frac{{{a_{2n + 1}} - {a_1}}}{{{a_{2n + 1}} + {a_1}}} + \frac{{{a_{2n + 1}} - {a_2}}}{{{a_{2n + 1}} + {a_2}}} + ...... + \frac{{{a_{2n + 1}} - {a_n}}}{{{a_{2n + 1}} + {a_n}}}$$

A
$$\frac{{n\left( {n + 1} \right)}}{2}\,\,\frac{{{a_2} - {a_1}}}{{{a_{n + 1}}}}$$

B
$$\frac{{n\left( {n + 1} \right)}}{2}$$

C
$$\left( {n + 1} \right)\left( {{a_2} - {a_1}} \right)$$

D
$$\frac{{n\left( {n + 1} \right)}}{2}\,\,\frac{{{a_{n + 1}} - {a_n}}}{{{a_n}}}$$

Solution

$${a}_{1},{a}_{2},...,{a}_{2n+1}$$ are in A.P
$$\Rightarrow\,{a}_{2n+1}={a}_{1}+2nd$$
$${a}_{2n}={a}_{1}+\left(2n-1\right)d$$
$$...{a}_{2}={a}_{1}+d$$
$$\Rightarrow\,{a}_{2n+1}-{a}_{1}=2{a}_{1}+2nd$$
$${a}_{2n}-{a}_{2}=2{a}_{1}+2nd$$ and so on
Consider $$\dfrac{{a}_{2n+1}-{a}_{1}}{{a}_{2n+1}+{a}_{1}}+\dfrac{{a}_{2n}-{a}_{2}}{{a}_{2n}+{a}_{2}}+\dfrac{{a}_{2n-1}-{a}_{3}}{{a}_{2n-1}+{a}_{3}}+...+\dfrac{{a}_{n+2}-{a}_{n}}{{a}_{n+2}+{a}_{n}}$$
$$=\dfrac{2nd}{2{a}_{1}+2nd}+\dfrac{\left(2n-2\right)d}{2{a}_{1}+2nd}+...+\dfrac{2d}{2{a}_{1}+2nd}$$
$$=\dfrac{nd}{{a}_{1}+nd}+\dfrac{\left(n-1\right)d}{{a}_{1}+nd}+...+\dfrac{d}{{a}_{1}+nd}$$
$$=\dfrac{d\left(1+2+3+...+n\right)}{{a}_{1}+nd}$$
$$=\dfrac{d\dfrac{n\left(n+1\right)}{2}}{{a}_{1}+nd}$$
$$=\dfrac{n\left(n+1\right)d}{2{a}_{n+1}}$$
$$=\dfrac{n\left(n+1\right)\left({a}_{2}-{a}_{1}\right)}{2{a}_{n+1}}$$
Question 8
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$$33!$$ is divisble by $${2^n}$$, then $$n$$=......

A
$$16$$

B
$$31$$

C
$$17$$

D
$$19$$

Solution

Given

$$33!$$

$$\implies 33\times 32 \times 31\times...\times 4\times 2\times 1$$

Considering the factors which contain 2

$$\implies 32\times 30\times 28\times 26\times 24\times 22\times 20\times 18 \times 16\times 14\times 12\times 10\times 8\times 6\times 4\times 2$$

$$\implies 2^5\times 2^1 \times 2^2 \times 2^1 \times 2^3 \times 2^1 \times 2^2 \times 2^1 \times 2^4 \times 2 \times 2^2 \times 2^1 \times 2^3 \times 2^1 \times 2^2 \times 2^1$$

$$\implies 2^{31}$$

$$\therefore n=31$$
Question 9
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Let $$P(n)$$ be the statement $$2^{n}<n!$$ where $$n$$ is a natural number, then $$P(n)$$ is true for:

A
all $$n$$

B
all $$n>2$$

C
all $$n>3$$

D
all $$n<3$$

Solution

We have,

$$P\left( n \right)$$ be the statement $${{2}^{n}}<n!$$

Where $$n$$ is a natural number

Then,

Put $$n=1,2,3....$$

So,

$$P\left( 1 \right)\,$$ be the statement of $${{2}^{1}}<1!=2<1\,\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$

$$P\left( 2 \right)$$ be the statement of $${{2}^{2}}<2!=4<2\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$
$$P\left( 3 \right)$$ be the statement of $${{2}^{3}}<3!=8<6\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$

But,
$$P\left( 4 \right)$$ be the statement of $${{2}^{4}}<4!=16<24\,\,\left( \text{It}\,\text{is}\,\text{right} \right)$$

Similarly,

$$P\left( 5 \right),\,P\left( 6 \right)\,.......$$ So, all number $$n$$ is a natural number.

Therefore, it is true for all $$n>3$$.

Hence, this is the answer.
Question 10
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For all positive integrals $$10^{n}+3^{4n+2}+8$$ is divisible by

A
$$8$$

B
$$9$$

C
$$7$$

D
$$6$$

Solution

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Principle of Mathematical Induction Test - 18

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Principle of Mathematical Induction Test - 18

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Question 1

$$a+b = c+d$$

$$a-b=c-d$$

$$a+b=c+d, a^2+b^2=c^2+d^2$$

$$a-b=c-d, a^2-b^2=c^2-d^2$$

Solution

Question 2

$$3$$

$$7$$

$$5$$

$$6$$

Solution

Question 3

7

9

1

3

Solution

Question 4

$$\dfrac { n\left( n+1 \right) \left( n+2 \right) }{ 6 } $$

$$\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } $$

$$\dfrac { n\left( n+1 \right) \left( 2n+7 \right) }{ 6 } $$

$$\dfrac { n\left( n+1 \right) \left( 2n+9 \right) }{ 6 } $$

Solution

Question 5

$$\begin{pmatrix}1 & n & n^{2}\\ 0 & n^{2} & n\\ 0 & 0 & n\end{pmatrix}$$

$$\begin{pmatrix}1 & n & n\left (\dfrac {n + 1}{2}\right )\\ 0 & 1 & n\\ 0 & 0 & 1\end{pmatrix}$$

$$\begin{pmatrix}1 & n^{2} & n\\ 0 & n & n^{2}\\ 0 & 0 & n^{2}\end{pmatrix}$$

$$\begin{pmatrix}1 & n & 2n - 1\\ 0 & \dfrac {n + 1}{2} & n^{2}\\ 0 & 0 & \dfrac {n + 1}{2}\end{pmatrix}$$

Solution

Question 6

$$2^{n+1}$$ as a factor

$$2^{n+2}$$ as a factor

$$2^{n+3}$$ as a factor

$$2^{n}$$ as a factor

Solution

Question 7

$$\frac{{n\left( {n + 1} \right)}}{2}\,\,\frac{{{a_2} - {a_1}}}{{{a_{n + 1}}}}$$

$$\frac{{n\left( {n + 1} \right)}}{2}$$

$$\left( {n + 1} \right)\left( {{a_2} - {a_1}} \right)$$

$$\frac{{n\left( {n + 1} \right)}}{2}\,\,\frac{{{a_{n + 1}} - {a_n}}}{{{a_n}}}$$

Solution

Question 8

$$16$$

$$31$$

$$17$$

$$19$$

Solution

Question 9

all $$n$$

all $$n>2$$

all $$n>3$$

all $$n<3$$

Solution

Question 10

$$8$$

$$9$$

$$7$$

$$6$$

Solution

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