MCQ Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 5

The integer next above

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

0%
$2^{n+1}$ as a factor
0%
$2^{n+2}$ as a factor
0%
$2^{n+3}$ as a factor
0%
$2^{n}$ as a factor

Explanation

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.

${a_{1,}}{a_{2,}}{a_{3,}}........{a_{2n + 1}}$ are in A.P. then

$\frac{{{a_{2n + 1}} - {a_1}}}{{{a_{2n + 1}} + {a_1}}} + \frac{{{a_{2n + 1}} - {a_2}}}{{{a_{2n + 1}} + {a_2}}} + ...... + \frac{{{a_{2n + 1}} - {a_n}}}{{{a_{2n + 1}} + {a_n}}}$

0%
$\frac{{n\left( {n + 1} \right)}}{2}\,\,\frac{{{a_2} - {a_1}}}{{{a_{n + 1}}}}$
0%
$\frac{{n\left( {n + 1} \right)}}{2}$
0%
$\left( {n + 1} \right)\left( {{a_2} - {a_1}} \right)$
0%
$\frac{{n\left( {n + 1} \right)}}{2}\,\,\frac{{{a_{n + 1}} - {a_n}}}{{{a_n}}}$

Explanation

${a}_{1},{a}_{2},...,{a}_{2n+1}$ 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

$\dfrac{{a}_{2n+1}-{a}_{1}}{{a}_{2n+1}+{a}_{1}}+\dfrac{{a}_{2n}-{a}_{2}}{{a}_{2n}+{a}_{2}}+\dfrac{{a}_{2n-1}-{a}_{3}}{{a}_{2n-1}+{a}_{3}}+...+\dfrac{{a}_{n+2}-{a}_{n}}{{a}_{n+2}+{a}_{n}}$

$=\dfrac{2nd}{2{a}_{1}+2nd}+\dfrac{\left(2n-2\right)d}{2{a}_{1}+2nd}+...+\dfrac{2d}{2{a}_{1}+2nd}$

$=\dfrac{nd}{{a}_{1}+nd}+\dfrac{\left(n-1\right)d}{{a}_{1}+nd}+...+\dfrac{d}{{a}_{1}+nd}$

$=\dfrac{d\left(1+2+3+...+n\right)}{{a}_{1}+nd}$

$=\dfrac{d\dfrac{n\left(n+1\right)}{2}}{{a}_{1}+nd}$

$=\dfrac{n\left(n+1\right)d}{2{a}_{n+1}}$

$=\dfrac{n\left(n+1\right)\left({a}_{2}-{a}_{1}\right)}{2{a}_{n+1}}$

State the whether given statement is true or false

Prove that by P.M.I.

$1^2+3^2+........+(2n-1)^2=\frac{n(4n^2-1)}{3}$

0%
True
0%
False

Explanation

$1^2+3^2+.....(2_{n-1})^2=\dfrac {n(4n^2-1)}{3}$

put

$n=1$

$LHS=1^2=1$

$RHS=\dfrac {1(4(1)^2-1)}{2}=1$

$LHS=RHS$

$\Rightarrow \ n=1$ statics the equation

put

$n=2$

$LHS=1^2+3^2=10$

$RHS=\dfrac {2(4(2)^2-1)}{3}=\dfrac {2\times 5}{3}=10$

$LHS=RHS$

$\Rightarrow \ n=2$ satisties the equation

Lets answer

$n=l'$ satisties the equation

$1^2+3^2+......(2l-1)^2=\dfrac {l(4l^2-1)}{3}$

add

$(2l+1)^2$ on both sides

$1^2+3^2+,,,,(2l-1)^2=\dfrac {l(4l^2-1)}{3}+(2l-1)^2$

$=\dfrac {4l^3-l+3(4l^2+1+4l)}{3}$

$=\dfrac {4l^3+12^2+3+11l}{3}$

$=\dfrac {(l+1)(4l^2+8l+3)}{3}$

$=(l+1)(4(l^2+2l-1)-1)$

$1^2+3^2+....(2l+1)^2=\dfrac {(l+1)(4(l+1)^2-1)}{3}$

$\therefore \ n=(l+1)$ also true

$\therefore$ its true fot all

$\in N$

16 divides

${n^4} + 4{n^2} + 11$ , if n is an odd integer.

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True
0%
False

The product of two consecutive positive integers is divisible by

0%
True
0%
False

$^{\text{1}}{{\text{p}}_{\text{1}}}{\text{ + 2}}{{\text{.}}^{\text{2}}}{{\text{p}}_{\text{2}}}{\text{ + 3}}{{\text{.}}^{\text{3}}}{{\text{p}}_{\text{3}}}{\text{ + }}....{\text{ + n}}{{\text{.}}^{\text{n}}}{{\text{p}}_{\text{n}}}{{\text{ = }}^{{\text{n + 1}}}}{{\text{p}}_{{\text{n + 1}}}}{\text{ -2}}$

0%
True
0%
False

$33!$ is divisble by

${2^n}$ , then

$n$ =......

0%
$16$
0%
$31$
0%
$17$
0%
$19$

Explanation

Given

$33!$

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

Considering the factors which contain 2

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

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

$\implies 2^{31}$

$\therefore n=31$

By induction it can be proved

$n ^ { 7 } - 7 n ^ { 5 } + 14 n ^ { 3 } - 8 n$ is divisible by 8.

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True
0%
False

Using P. M. 1, whether

$41^n-14^n$ is a multiple of

$27$ .

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True
0%
False

Use Principle of Mathematical Induction it can be proved

$1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + \cdots + n ^ { 3 } = \left( \frac { n ( n + 1 ) } { 2 } \right) ^ { 2 }$ for all

$n \in N$ .

0%
True
0%
False

Prove by method of induction, for all

$n \in$

$3 + 7 + 11 + ...$ to

$n$ terms

$= n(2n + 1)$

State True or False

0%
True
0%
False

Explanation

Given statement

$P(n)$ :

$3+7+11+....$ to

$n$ terms

$=n(2n+1)$

$\begin{array}{l} Put\, \, n=1 \\ L.HS=3 \\ R.H.S=1.\left( { 2.1+1 } \right) =3 \\ L.HS=R.HS \\ P\left( 1 \right) \, \, is\, \, true. \end{array}$

Assuming

$P(k)$ is true, where

$k \in N$

$P(k):3+7+.....$ to

$k$ terms

$=k(2k+1)...............(1)$

Consider

$P(k+1)$

$L.H.S=3+7+............$ to

$k+1$ term

$=3+7+.......$ to

$k$ terms +

$(4k+3)$

Where

$4k+3$ is

$(k+1)$ th term of given series.

Using in

$L.H.S$ we get

$\begin{array}{l} L.H.S=k\left( { 2k+1 } \right) +4k+3 \\ =2{ k^{ 2 } }+5k+3 \\ =\left( { k+1 } \right) \left( { 2k+3 } \right) \\ =\left( { k+1 } \right) \left[ { 2\left( { k+1 } \right) +1 } \right] \end{array}$

Hence,

$P(k+1)$ is true.

Let

$P(n)$ be the statement

$2^{n}<n!$ where

$n$ is a natural number, then

$P(n)$ is true for:

0%
all $n$
0%
all $n>2$
0%
all $n>3$
0%
all $n<3$

Explanation

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.

Using principle of mathematic inductor for all

$n\ \epsilon\ N:1+2+3+....+n < \dfrac {1}{8}(2n+1)^{2}$

0%
True
0%
False

$\sin(\alpha )+\sin\left( \alpha +\dfrac { \pi }{ 6 } \right) +\sin\left( \alpha +\dfrac { 2\pi }{ 6 } \right) +...$

$+\sin\left( \alpha +\dfrac { (n-1)\pi }{ 6 } \right)=\frac { \sin\left( \alpha +\frac { (n-1)\pi }{ 6 } \right) \times \sin\left( \dfrac { n\pi }{ 12 } \right) }{ \sin\left( \dfrac { \pi }{ 12 } \right) }$

0%
True
0%
False

Using mathematical induction, for all

$n\epsilon N$

${ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ n }^{ 2 }=\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }$

0%
True
0%
False

By method of Induction for all

$n\in N$

$\dfrac { 1 }{ 3.3 } +\dfrac { 1 }{ 3.5 } +\dfrac { 1 }{ 5.7 } +....+\dfrac { 1 }{ (2n-1)(2n+1) } =\dfrac { n }{ 2n+1 }$

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True
0%
False

Consider the statement:

$"P(n):n^2-n+41$ is prime". Then which one of the following is true?

0%
P $(5)$ is false but P $(3)$ is true
0%
Both P $(3)$ and P $(5)$ are false
0%
P $(3)$ is false but P $(5)$ is true
0%
Both P $(3)$ and P $(5)$ are true

Explanation

$P(n): n^2-n+41$ is prime

$P(5)=61$ which is prime

$P(3)=47$ which is also prime.

For all positive integrals

$10^{n}+3^{4n+2}+8$ is divisible by

0%
$8$
0%
$9$
0%
$7$
0%
$6$

$n(n^2+5)$ divisible by

$6$ ,

$\forall n\in Z$ ?

0%
True
0%
False

Explanation

Let,

$P(n) = n(n^2 + 5)$ be divisible by 6 for all integer values of '

$n$ '.

Substituting

$n = 1$ , we get,

$\Rightarrow P(1) = 1(1^2 + 5) = 6$ , which is divisible by 6.

$\therefore P(1)$ is true.

Let,

$n = k$ for any integer '

$k$ '.

$P(k) = k (k^2 + 5)$ , is divisible by 6. ....(1)

$\Rightarrow P(k+1) = (k + 1) [ (k+1)^2 + 5]$

$=(k+1)(k^2 + 2k+6)$

$=k^3 + 2k^2 +6k +k^2 + 2k + 6$

$=k^3 + 3k^2 + 8k + 6$

$=k^3+5k + 3k^2 + 3k + 6$

$=k(k^2 + 5)+3(k^2+k+2)$

$=k(k^2 + 5) + 3 [k(k+1)+2]$

$\therefore$ The 1st part is divisible by 6 [from (1)] and 2nd part is also divisible by 3. As it will be an even number for all integer values of '

$k$ ', since either

$k$ or

$k+1$ is always an even number, thus adding 2 to it yields another even number, thus 2nd part is also divisible by 6.

Hence, option (A) is the correct answer.

$4^n-1$ is divisible by

$3$ ?

0%
True
0%
False

$2n<(n+2)!$ for all natural number

$n?$

0%
True
0%
False

$n^3-7n+3$ divisible by

$3$ ?

0%
True
0%
False

Explanation

Let

$P(n)=n^3-7n+3$
When

$n=1$

$P(1)= 1-7+3 = -3$ which is divisible by 3.
Assume

$P(k)$ is divisible by 3.
Then

$k^3-7k+3 = 3M..........(1)$
Consider

$P(k+1)$

$P(k+1)=(k+1)^3-7(k+1)+3$

$=k^3+3k^2+3k+1-7k-7+3$

$=k^3+3k^2-4k-3$
From (1)

$P(k+1)= 3M+7k-3+3k^2-4k-3$

$=3(M+k^2-2)$
which is divisible by 3.

$P(k+1)$ is true.
By principle of mathematical induction

$n^3-7n+3$ is always divisible by 3.

$2+4+6+......2n=n^2+n$ true for all natural numbers

$n$ ?

0%
True
0%
False

$n^2<2^n$ for all natural numbers

$n\geq 5?$

0%
True
0%
False

$3^{2n}-1$ is divisible by

$8?$

0%
True
0%
False

State true or false.

$1+5+9+........+(4n-3)=n(2n-1)$ for all natural numbers

$n$ ?

0%
True
0%
False

Explanation

Given,

$1+5+9+...+(4n-3)=n(2n-1)$

Let

$P(n) = 1 + 5 + 9 + ... + (4n-3) = n(2n-1)$ for all natural numbers

Let

$n=1$

$P(1) = 1 (2\times 1)-1)$

$P(1) = 1 (2-1) \Rightarrow P(1) = 1\times 1 \Rightarrow P(1) =1$

Hence,

$P(1)$ is true

Assume that

$P(n)$ is true for some natural no.

$n = k$ .

$P(k) = 1 +5 + 9 + ... + (4k-3) = k(2k-1)$ ...(1)

We should prove that

$P(k+1)$ is true.

The

$n^{th}$ term of

$P(n) = 4n-3$ .

Next term:

$P(n+1) = 4(n+1)-3$ .

$=4n+4-3 = 4n+1$ .

Adding

$4k+1$ on both sides in equation (1).

$P(k+1) = 1+5+9+....+(4k-3)+(4k+1)=k(2k-1)+4k+1$

$=2k^2 - k + 4 k + 1$

$=2k^2 + 3k + 1$

$=2k^2+2k+k+1$

$=2k(k+1)+1(k+1)$

$=(k+1)(2k+1)$

$=(k+1)(2k+2-2+1)$

$=(k+1)(2(k+1)-1)$

$P(k+1)$ is true when

$P(k)$ is true.

Hence, by the principle of mathematical induction,

$P(n)$ is true for any natural number

$n$ .

Let P(n) be a statement and let

$P(k) \Rightarrow P(k + 1),$ for some natural number

$k,$ then

$P(n)$ is true for all

$n \in N.$

0%
True
0%
False

State true or false.

For any natural number n,

$7^n – 2^n$ is divisible by

$5.$

0%
True
0%
False

$1+2+2^2+......+2^n=2^{n+1}-1$ for all natural numbers n?

0%
True
0%
False

By mathamatical induction

$n(n^2 - 1)$ is divisible by

0%
$19$
0%
$23$
0%
$24$
0%
$29$

Explanation

Let

$P(n) = n(n^2 - 1)$

Step I: For

$n = 1$ ,

$P(1) = 1 (1^1 - 1)$

$=$ 0, 0 is divisible by 24.

Therefore, the result is true for

$n=1$ .

Step II: Assume that the result is true for

$n = k$
i.e.,

$P(k) = k (k^2 -1)$ is divisible by 24

$\Rightarrow P(k) = 24r$ , when r is an integer and k is an odd positive integer.

Step III: For

$n = k + 2$

$\therefore P(k + 2) = (k +2)((k + 2)^2 -1)$

$= (k + 2) (k^2 + 4k + 3)$

$= k^3 + 6k^2 + 11k + 6$

$= 24 r + k + 6k^2 + 11k + 6$ (From assumption step)

$= 6k^2 + 12k + 6 + 24 r$

$= 6(k + 1)^2 + 24 r$

$= 6(2m)^2 + 24 r$ (

$\because$ k + 1 is even, let k + 1

$=$ 2m)

$= 24 (m^2 + r)$

$\because m^2 + r$ is an integer,

$\therefore 24(m^2 + r)$ is clearly divisible by 24.

Hence

$P(k + 2)$ is divisible by

$24$ . This shows that the result is true for

$n = k + 2$ .
Hence by the principle of mathematical induction, the result is true for n odd integer.

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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