MCQ 511 Mark
$\ce{x(x^{n-1}- nα^{n-1}) + α^n(n - 1)}$ is divisible by $\ce{(x - α)^2}$ for:
- A
$n > 1$
- B
$n > 2$
- ✓
all $n \in N$
- D
AnswerCorrect option: C. all $n \in N$
View full question & answer→MCQ 521 Mark
Let $T(k)$ be the statement $1 + 3 + 5 + .... + (2k – 1)= k +10$ Which of the following is correct:
AnswerCorrect option: B. $T(k)$ is true $\Rightarrow T(k + 1)$ is true
View full question & answer→MCQ 531 Mark
The number of values of $n,$ for which $p(n) = 1! + 2! + 3! + 4!+ ...... + n!$ is the square of a natural number, is equal to:
AnswerFor $n = 4, P(n) = 1 + 2 + 6 + 24 = 33.$
For $n > 4, n!$ will always have $0$ in the units place.
$3$ will be unit place digit of $p(n)$
hence can not be sqaure of any natural no.
So the $n = 1, 3$ are the answer.
View full question & answer→MCQ 541 Mark
If $n \in N,$ then $11^{n+2}+ 12^{2n+1}$ is divisible by:
View full question & answer→MCQ 551 Mark
$1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1):$
- A
$n(n + 1)(n + 2)$
- B
$\{n(n + 1)(n + 2)\}/2$
- ✓
$\{n(n + 1)(n + 2)\}/3$
- D
$\{n(n + 1)(n + 2)\}/4$
AnswerCorrect option: C. $\{n(n + 1)(n + 2)\}/3$
Let the given statement be $P(n).$ Then,
$P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = (1/3)\{n(n + 1) (n + 2)\}$
Thus, the given statement is true for $n = 1,$
i.e., $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + k(k + 1) = (1/3)\{k(k + 1) (k + 2)\}.$
Now, $1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +…+ k(k + 1) + (k + 1) (k + 2)$
$= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ……. + k(k + 1)) + (k + 1) (k + 2)$
$= (1/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [$using $(i)]$
$= (1/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)$
$= (1/3)\{(k + 1) (k + 2)(k + 3)\}$
$\Rightarrow P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)$
$= (1/3)\{k + 1 )(k + 2) (k +3)\}$
$\Rightarrow P(k + 1)$ is true, whenever $P(k)$ is true.
Thus, $P(1)$ is true and $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all values of $\in N$
View full question & answer→MCQ 561 Mark
$\forall n \in N; x^{2n-1}+ y^{2n-1}$ is divisible by?
- A
$x − y$
- ✓
$x + y$
- C
$xy$
- D
$x^2+ y^2$
AnswerCorrect option: B. $x + y$
View full question & answer→MCQ 571 Mark
If $P(n): “46 + 16 + k$ is divisible by $64$ for $\ce{n Î N”}$ is true, then the least negative integral value of $k$ is:
View full question & answer→MCQ 581 Mark
For all $10^n+ 3(4^{n+2}) + 5$ is divisible by $(n \in N):$
View full question & answer→MCQ 591 Mark
If $n \in N, 7^{2n}48^{n-1}–$ is divisible by:
AnswerCorrect option: D. $2304$
View full question & answer→MCQ 601 Mark
Let $x > -1,$ then statement $p(n) : (1+x)^n > 1 + nx $, where $n \in N$ is true for
AnswerCorrect option: C. For all $n > 1,$ provided $x = 0.$
$p(1):(1+x)^1>1+x$ is false
$p(2):(1+x)^2>1+2 x \Rightarrow x^2>0$ is true when $x=0$
$p(3):(1+x)^3>1+3 x \Rightarrow x^2>0$ is true when $x=0$
Let $p(k)(1+x)^k>1+k x$ is true for some $k \in N, k>1$
$ \Rightarrow(1+x)^{k+1}>(1+k x)(1+x)$
$ \Rightarrow(1+x)^{k+1}>1+(k+1) x+k x^2>1+(k+1) x, x=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 $x = 0$
View full question & answer→MCQ 611 Mark
$P(n): 2 \times 7n + 3 \times 5n - 5$ is divisible by:
- ✓
$24, \forall n \in N$
- B
$21, \forall n \in N$
- C
$32, \forall n \in N$
- D
$50, \forall n \in N$
AnswerCorrect option: A. $24, \forall n \in N$
Concepts:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$
i.e., the truth of $P(k)$ implies the truth of $P(k + 1).$
Then, $P(n)$ is true for all natural numbers $n$
Calculation:
Given:
$P(n) = 2 \times 7n + 3 \times 5n - 5$
Put $n = 1$
$P(1) = 2 \times 7^1+ 3 \times 5^1- 5 = 24$, Which is divisible by $24$
Assume $P(k)$ is true
$P(k) = 2 \times 7^k+ 3 \times 5^k- 5 = 24q$, where $q\ \epsilon\ N .....(1)$
Now,
$ T(k+1)=2 \times 7^{k+1}+3 \times 5^{k+1}-5=2 \times 7^k \times 7+3 \times 5^k \times 5-5 $
$ \Rightarrow 7\left\{2 \times 7^k+3 \times 5^k-5-3 \times 5^k+5\right\}+3 \times 5^k \times 5-5 $
$ \Rightarrow 7\left\{24 q-3 \times 5^k+5\right\}+15 \times 5^k-5 $
$ \Rightarrow(7 \times 24 q)-21 \times 5^k+35+15 \times 5^k-5 $
$ \Rightarrow(7 \times 24 q)-6 \times 5^k+30=(7 \times 24 q)-6\left(5^k-5\right) $
$\Rightarrow(7 \times 24 q)-6 \times(4 p)$ As $(5^k-5) $ is a multiple of $4 \}$
$\Rightarrow (7 \times 24q) - 24p = 24(7q - p)$
$\Rightarrow 24 \times r, r = 7q - p,$ is some natural number $.......(2)$
Thus, $P(k + 1)$ is true whenever $P(k)$ is true
View full question & answer→MCQ 621 Mark
For each natural number, the statement $P(n) = 2^{3n}− 1$ is divisible by:
Answer$ P(n)=2^{3 n}-1=8^n-1 $
$ P(n)=(1+7)^n-1 $
$ \Rightarrow P(n)=1+{ }^n C_1 \times 7+{ }^n C_2 \times 7^2+\ldots+{ }^n C_n \times 7^n-1 $
$ \Rightarrow P(n)=7\left({ }^n C_1+{ }^n C_2 7+\ldots+{ }^n C_n 7{ }^{n-1}\right)$
Therefore, $P(n)$ is divisible by $7.$
View full question & answer→MCQ 631 Mark
Consider the following statements : $1) \sim (p\ ∧ q) = \sim p\ ∨ \sim q 2) \sim (p\ ∨ q) = \sim p\ ∧ \sim q 3) \sim (\sim p) = p$ Which of the above statements is/are correct?
- A
$1$ and $2$
- B
$2$ and $3$
- ✓
$1, 2$ and $3$
- D
AnswerCorrect option: C. $1, 2$ and $3$
Concept:
The negation of a conjunction $p ∧ q$ is the disjunction of the negation of $p$ and the negation of $q.$ Equivalently, we write $\sim (p ∧ q) = \sim p ∨ \sim q$
The negation of a disjunction $p ∨ q$ is the conjunction of the negation of $p$ and the negation of $q.$ Equivalently, we write $\sim (p ∨ q) = \sim p ∧ \sim q$
Negation of negation of a statement is the statement itself. Equivalently, we write $\sim (\sim p) = p$
View full question & answer→MCQ 641 Mark
For each $n\ \epsilon\ N,$ then $3^{2n+1}+ 1$ is divisible by:
View full question & answer→MCQ 651 Mark
If $n \in N \Big(\frac{\text{n}+1}{2}\Big)^\text{n}\geq\text{n!}$ is true when:
- ✓
$n ≥ 1$
- B
$n ≥ 2$
- C
$n > 1$
- D
$n > 2$
AnswerCorrect option: A. $n ≥ 1$
Concept:
$\text{n!}=\text{n}\times(\text{n}-1)\times\text{n}-2.....\times3\times2\times1$
Calculation:
Given:
$\text{p}\text{(n)}=\Big(\frac{\text{n+1}}{2}\Big)^\text{n}\geq\text{n!}$
Put $n = 1$
$\text{p}(2)=\Big(\frac{2+1}{2}\Big)^2\geq2!$
$=\Big(\frac{3}{2}\Big)^2\geq2\times1$
$2.25 ≥ 2$
Put $n = 3$
$\text{p}(3)=\Big(\frac{3+1}{2}\Big)^3\geq3!$
$8\geq3\times2\times1,8\geq6$
Hence, The given expression $P(n)$ is true for $n ≥ 1$
View full question & answer→MCQ 661 Mark
For all positive integral values ofn, $3^{2n} - 2n + 1$ is divisible by:
View full question & answer→MCQ 671 Mark
If $P(n)$ be the statement $n(n + 1) + 1$ is odd, then which of the following is false?
Answer$P(n) = n(n + 1) + 1$
$P(2) = 6 + 1 = 7$
$P(3) = 3 \times 4 + 1 = 13$
$P(4) = 4 \times 5 + 1 = 21$
None of the above is even.
View full question & answer→MCQ 681 Mark
$n(n + 1)(n + 5)$ is a multiple of $3$ is true for:
AnswerCorrect option: C. All natural numbers $n$
View full question & answer→MCQ 691 Mark
If $P(n): “49^n+ 16^n+ k$ is divisible by $64$ for $n \in N”$ is true, then the least negative integral value of $k$ is:
AnswerGiven that $P(n): 49^n+ 16^n+ k$ is divisible by $64$ for $n \in N$
For $n = 1,$
$P(1): 49 + 16 + k = 65 + k$ is divisible by $64.$
Thus $k,$ should be $-1$ since, $65 - 1 = 64$ is divisible by $64.$
View full question & answer→MCQ 701 Mark
For all $n \in N, 3.5^{2n+1}+ 2^{3n+1}$ is divisible by:
View full question & answer→MCQ 711 Mark
If $n$ is an odd positive integer, then $a^n + b^n$ is divisible by:
- A
$a^2 + b^2$
- ✓
$a + b$
- C
$a – b$
- D
AnswerCorrect option: B. $a + b$
Given number $= a^n + b^n$
Let $n = 1, 3, 5, ……..$
$a^n + b^n = a + b$
$a^n + b^n = a^3 + b^3 = (a + b) \times (a^2 + b^2 + ab)$ and so on.
Since, all these numbers are divisible by $(a + b)$ for $n = 1, 3, 5,…..$
So, the given number is divisible by $(a + b)$
View full question & answer→MCQ 721 Mark
For all $n ≥ 2,n^2(n^4-1)$ is divisible by:
View full question & answer→MCQ 731 Mark
If $10^\text{n} + 3 \times 4^{\text{n}+2}+\lambda$ is divisible by $9$ for all $\text{n}\in\text{N},$ then the least positive integer value of $\lambda$ is
AnswerGiven,
$10\text{n }+3\times^{\text{n}+2}+\lambda$ is divisible by $9,$
$\text{P}(1)=10^1+3\times4^{1+2}+\lambda$ is exactly divisible by $9$ then the value of $\lambda$ is $5.$
View full question & answer→MCQ 741 Mark
For all $n\in N, 72n − 48n−1$ is divisible by:
AnswerCorrect option: B. $2304$
Concepts:
Suppose there is a given statement $P (n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P (1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$ i.e., truth of $P (k)$ implies the truth of $P (k + 1).$
Then, $P (n)$ is true for all natural numbers $n$
Calculation:
Given:
$P(n) = 72n − 48n−1$
Put, $n = 1$
$P(1) = 72 − 48 \times 1 −1 = 0$
Check the expression $P(n)$ for $n = k ($where $k$ is some positive integer$) = 2, 3, 4......$
$P(2) = 7^{2n}− 48n − 1$
$= 7^4− 48 \times 2 − 1$
$= 2401 – 96 – 1$
$= 2401 – 97$
$= 2304$
$P(3) = 7^{2n}− 48n − 1$
$= 7^6− 48 \times 3 − 1$
$= 117649 – 144 – 1$
$= 117649 – 145$
$= 117504$
$= 2304 \times 51$
Since, all these numbers are divisible by $2304$ for $n = 1$ and $k = 2, 3,…..$
So, the given number is divisible by $2304$
View full question & answer→MCQ 751 Mark
Let $P(n)$ be the statement that $n^2 - n + 41$ is prime, then which of the following is not true?
AnswerCorrect option: C. $P(41)$
given that $P(n) = n^2- n + 41$
$P(2) = 2^2- 2 + 41 = 43 ($prime true$)$
$P(3) = 3^2- 3 + 41 = 47 ($prime true$)$
$P(41) = 41^2- 41 + 41 = (41)^2$
$\therefore P(41 )$ is not true.
View full question & answer→MCQ 761 Mark
What is the sum of $2 + 4 + 6 + 8 +....+ 2n\ ?\ A:$
- ✓
$n(n + 1)$
- B
$n(n + 2)$
- C
$n(n + 3)$
- D
$n(n + 4)$
AnswerCorrect option: A. $n(n + 1)$
View full question & answer→MCQ 771 Mark
If $x^n− 1$ is divisible by $x − k,$ then the least positive integral value of $k$ is:
Answerif $f(x) = x^n− 1$ is divisible by $x − k$
Then $f(k) = 0$
Therefore, $k^n= 1$
and thus least positive integral value of $k$ is $1$
View full question & answer→MCQ 781 Mark
The smallest positive integer $n$ for which $ \text{n}!<\Big(\frac{\text{n}+1}{2}\Big)\text{n}$ holds, is:
View full question & answer→MCQ 791 Mark
$72n + 16n - 1$ is divisible by $(n \in N):$
Answer$S = 72n + 16n - 1$
For $n = 1$
$S = 49 + 16 - 1 = 64$
For $n = 2$
$S = 2401 + 32 - 1$
$= 2432$
$= 64 \times 32,$
which is divisible by $64$
View full question & answer→MCQ 801 Mark
What would be the hypothesis of mathematical induction for $n(n + 1) < n! ($where $n ≥ 4)\ ?$
- A
It is assumed that at $n = k, k(k + 1)! > k!$
- ✓
It is assumed that at $n = k, k(k + 1)! < k!$
- C
It is assumed that at $n = k, k(k + 1)! > (k + 1)!$
- D
It is assumed that at $n = k, (k + 1)(k + 2)! < k!$
AnswerCorrect option: B. It is assumed that at $n = k, k(k + 1)! < k!$
When we use the principle of mathematical induction, we assume that $P(n)$ is true for $P(k)$ and prove that $P(k + 1)$ is also true.
Here $P(k)$ is $k(k + 1)! < k!$
View full question & answer→MCQ 811 Mark
If $P(n): 3n < n!, n\ \epsilon\ N,$ Then $P(n)$ is true for:
- ✓
$n ≥ 7$
- B
$n ≥ 3$
- C
$n ≥ 6$
- D
AnswerCorrect option: A. $n ≥ 7$
Concept:
$\text{n}!=\text{n}\times(\text{n}-1)\times(\text{n}-2).....\times3\times2\times1$
Given:
$P(n): 3n < n!$
This can be solved directly by hit and trial method, putting the option in expression and checking its validity
We will choose first the smallest number from the options
Putting $n = 3$
$ P(n) = 3n < n! = 3^3 < 3! $
$\Rightarrow 27 ≮ 3 \times 2 \times 1, 27 ≮ 6 ,$
Hence Option $2$ is wrong
Putting $n = 6$
$P(n) = 3n < n! = 63 < 6!$
$P(n) = 3n < n! = 3^6 < 6! , 1029 < 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$1029 ≮ 720$ Hence Option $3$ is wrong
Put $n = 7$
$P(n) = 3n < n! = 3^7 < 7! , 2187 < 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$2187 < 5040,$ Option $1$ satisfies the given expression,
Hence the correct answer is option $1.$
View full question & answer→MCQ 821 Mark
For all $n\in N, 5^{2n}− 1$ is divisible by:
AnswerGiven number $= 5^{2n}− 1$
Let $n = 1, 2, 3, 4, ……..$
$ 5^{2 n}-1=5^2-1=25-1=24 $
$ 5^{2 n}-1=5^4-1=625-1=624=24 \times 26 $
$ 5^{2 n}-1=5^6-1=15625-1=15624=651 \times 24 $
Since, all these numbers are divisible by $24$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $2$
View full question & answer→MCQ 831 Mark
For a positive integer $n,$ let $\text{a(n)} =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…+\frac{1}{\text{n.}} $ Then,
- ✓
$a(100) ≤ 100$
- B
$a(100) >100$
- C
$a(200) ≤ 100$
- D
AnswerCorrect option: A. $a(100) ≤ 100$
View full question & answer→MCQ 841 Mark
A student was asked to prove a statement $P(n)$ by induction. He proved that $P(k + 1)$ is true whenever $P(k)$ is true for all $k > 5 \in N$ and also that $P(5)$ is true. Based on this, he could conclude that $P(n)$ is true:
- A
for all $n \in N$
- B
for all $n > 5$
- ✓
for all $n ≥ 5$
- D
for all $n < 5$
AnswerCorrect option: C. for all $n ≥ 5$
The student could be able to conclude that $P(n)$ is true for all $n ≥ 5$ since $P(5)$ is true for all $k > 5 \in N$ as well as true for $P(5)$ and $P(k + 1)$ is true, whenever $P(k)$ is true.
View full question & answer→MCQ 851 Mark
For positive integer $n, 10^{n-2} > 81 n,$ if:
- A
$n > 5$
- ✓
$n ≥ 5$
- C
$n < 5$
- D
$n > 6$
AnswerCorrect option: B. $n ≥ 5$
View full question & answer→MCQ 861 Mark
Choose the correct answer. If $x^n- 1$ is divisible by $x - k,$ then the least positive integral value of $k$ is:
AnswerLet $P(n) = x^n- 1$ is divisible by $x - k$
$P(1) = x - 1$ is divisible by $x - k.$
Since $x - 1$ is divisible by $x - 1,$ the least integral value of $k$ is $1.$
View full question & answer→MCQ 871 Mark
$n (n+1) (n+5)$ is a multiple of $3$ is true for
AnswerCorrect option: C. All natural numbers $n$
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) $
$k^3+ 9k^2+ 20k + 12 = (k^3+ 6k^2+ 5k) + (3k^2+15k+12)$
$= 3m + 3k^2+ 15k + 12 [$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 $n \in N$
View full question & answer→MCQ 881 Mark
The greatest positive integer, which divides $n(n +1)(n + 2)(n + 3)$ for all $\text{n Î N},$ is:
View full question & answer→MCQ 891 Mark
If $n \in N, 7^{2n}– 48n – 1$ is divisible by:
AnswerCorrect option: D. $2304$
View full question & answer→MCQ 901 Mark
If an $= \sqrt7 + \sqrt7 + \sqrt7 +......$ having n radical signs then by methods of mathematical induction which is true:
- A
$\text{an}>7"\text{n}\geq1$
- ✓
$\text{an}<7"\text{n}\geq1$
- C
$\text{an}<4"\text{n}\geq1$
- D
$\text{an}<3"\text{n}\geq1$
AnswerCorrect option: B. $\text{an}<7"\text{n}\geq1$
View full question & answer→MCQ 911 Mark
The sum of the series $1^3 + 2^3 + 3^3 + ………..n^3$ is:
- A
$\Big(\frac{\text{(n}+1)}{2}\Big)^2$
- B
$\Big(\frac{\text{n}}{2}\Big)^2$
- C
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)$
- ✓
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
AnswerCorrect option: D. $\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
Given, series is $1^3 + 2^3 + 3^3 + ……….. n^3$
$\Big(\frac{\text{n}\text{(n}+1)}{2}\Big)^2$
View full question & answer→MCQ 921 Mark
Choose the correct answer. For all $n \in N, 3 \times 5^{2n+1} + 2^{3n+1}$ is divisible by:
View full question & answer→MCQ 931 Mark
For $n\in N,\big(\frac{1}{5}\big)\text{n}^5+\big(\frac{1}{3}\big)\text{n}^3+\big(\frac{1}{15}\big)$ is:
View full question & answer→MCQ 941 Mark
$P(n) = n(n^2– 1)$. Which of the following does not divide $P(k+1)\ ?$
- A
$k$
- B
$k + 2$
- ✓
$k + 3$
- D
$k + 1$
AnswerCorrect option: C. $k + 3$
$P(n) = n(n^2- 1)$
$P(k + 1) = (k + 1) ((k + 1)^2-1)$
$P(k + 1) = (k + 1) (k^2+ 1 + 2k - 1)$
$P(k + 1) = (k + 1) (k^2+ 2k)$
$P(k + 1) = (k + 1) k (k + 2)$
Therefore, $k, (k + 1), (k - 1)$ divide $P(k + 1)$.
View full question & answer→MCQ 951 Mark
The greatest positive integer, which divides $(n + 2) (n + 3) (n + 4) (n + 5) (n + 6)$ for all $n \in N,$ is:
View full question & answer→MCQ 961 Mark
$n^2 < 2^n$ for all natural numbers:
- ✓
$n ≥ 5$
- B
$n < 5$
- C
$n > 1$
- D
$n ≤ 3$
AnswerCorrect option: A. $n ≥ 5$
Consider, $P(n) : n^2 < 2^n$
Substituting $n = 1, 2, 3,…$
$P(1): 1^2< 2^1$
$1 < 2 ($not true$)$
$P(2): 2^2 < 2^2$
$4 < 4 ($not true$)$
$P(3): 3^2 < 2^3$
$9 < 8 ($not true$)$
$P(4): 4^2 < 2^4$
$16 < 16 ($not true$)$
$P(5): 5^2 < 2^5$
$25 < 32 ($true$)$
$P(6): 6^2 < 2^6$
$26 < 64 ($true$)$
Thus, $n^2 < 2^n$ for all natural numbers $n ≥ 5.$
View full question & answer→MCQ 971 Mark
If $P(n)$ is a statement such that $P(3)$ is true. Assuming $P(k)$ is true $Þ P(k + 1)$ is true for all $\text{k} \geq 3$, then $P(n)$ is true:
- A
for all $n$
- ✓
for $n ≥ 3$
- C
for $n > 4$
- D
AnswerCorrect option: B. for $n ≥ 3$
View full question & answer→MCQ 981 Mark
Let $P(n)$ be a statement and $P(n) = P(n + 1)\forall n \in N,$ then $P(n)$ is true for what values of $n?$
AnswerCorrect option: A. For all $n$
Given, $P(n) = P(n+1)\forall n \in N$
Substituting $n - 1$ in place of $n,$
$P(n - 1) = P(n)$
Thus if $P(k)$ is true for some $k \in N,$ then it is true for $k - 1$ and $k + 1.$
Thus, it is true $\forall k \in N$
View full question & answer→MCQ 991 Mark
$x\left(x^{n-1}-n a^{n-1}\right)+a^n(n-1)$ is divisible by $(x-a)^2$ for:
- A
$n > 1$
- B
$n > 2$
- ✓
all $n \in N$
- D
AnswerCorrect option: C. all $n \in N$
View full question & answer→MCQ 1001 Mark
A student was asked to prove a statement $p(n)$ by induction. He proved $p(K + 1)$ is true whenever $p(k)$ is true for all $\text{k}>5\in\text{N}$ and also $p(5)$ is true. On the basis of this he could conclude that $p(n)$ is true.
AnswerCorrect option: C. For all $\text{n}\geq5$
$P(n)$ is true for all positive integer $n,$
i.e. $\text{n}\geq5,$
Where $P(n)$ is a Propositional function, complete two steps:
Basic Step: Verify that the proposition $P(1)$ is true.
Inductive Step: Show the conditional statement,
$\big[\text{P}(1) \wedge \text{P}(2) \wedge-\wedge\text{P}(\text{k})\big]\rightarrow\text{P(k+1)}$ holds for all positive integer.
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