MCQ 1511 Mark
Let $P(n) : n^2+ n + 1$ is an even integer. If $P(k)$ is assumed true $Þ P(k + 1)$ is true. Therefore, $P(n)$ is true:
- A
for $n > 1$
- B
for all $n \in N$
- C
for $n > 2$
- ✓
View full question & answer→MCQ 1521 Mark
$\{1/(3 ∙ 5)\} + \{1/(5 ∙ 7)\} + \{1/(7 ∙ 9)\} + ……. + 1/\{(2n + 1) (2n + 3)\}:$
- A
$n/(2n + 3)$
- B
$n/\{2(2n + 3)\}$
- ✓
$n/\{3(2n + 3)\}$
- D
$n/\{4(2n + 3)\}$
AnswerCorrect option: C. $n/\{3(2n + 3)\}$
Let the given statement be $P(n).$ Then,
$P(n): \{1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9)\} + ……. + 1/\{(2n + 1)(2n + 3)\} = n/\{3(2n + 3)\}.$
Putting $n = 1$ in the given statement, we get
and $\text{LHS} = 1/(3 ∙ 5) = 1/15$ and $\text{RHS} = 1/\{3(2 \times 1 + 3)\} = 1/15.$
$\text{LHS = RHS}$
Thus, $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): \{1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/\{(2k + 1)(2k + 3)\} = k/\{3(2k + 3)\} ….. (i)$
Now, $1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[\{2(k + 1) + 1\}2(k + 1) + 3$
$= \{1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]\} + 1/\{(2k + 3)(2k + 5)\}$
$= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [$using $(i)]$
$= \{k(2k + 5) + 3\}/\{3(2k + 3)(2k + 5)\}$
$= (2k^2 + 5k + 3)/[3(2k + 3)(2k + 5)]$
$= \{(k + 1)(2k + 3)\}/\{3(2k + 3)(2k + 5)\}$
$= (k + 1)/\{3(2k + 5)\}$
$= (k + 1)/[3\{2(k + 1) + 3\}]$
$= P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[\{2(k + 1) + 1\}\{2(k + 1) + 3\}]$
$= (k + 1)/\{3\{2(k + 1)\} + 3\}]$
$\Rightarrow P(k + 1)$ 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 $n \in N$
View full question & answer→MCQ 1531 Mark
Choose the correct answer. If $10^n+ 3 \times 4^{n+2} + k$ is divisible by $9$ for all $n \in N,$ then the least positive integral value of $k$ is:
AnswerLet $P(n) = 10^n+ 3 \times 4^{n+2}+ k$ is divisible by $9, \forall\text{ n}\in\text{N}$
$P(1) = 10^1+ 3 \times 4^{1+2}+ k$
$= 10 + 3 \times 64 + k$
$= 10 + 192 + k$
$= 202 + k$ must be divisible by $9.$
If $(202 + k)$ is divisible by $9$ then $k$ must be equal to $5.$
$202 + 5 = 207$ which is divisible by $9.$
$=\frac{207}{9}$
$=23$
So, the least positive integral value of $k = 5$
View full question & answer→MCQ 1541 Mark
For each $n \in N, 10^{2n-1}+ 1$ is divisible by
View full question & answer→MCQ 1551 Mark
For any natural number $n, 2^{2n} - 1$ is divisible by:
AnswerLet $P(n)=2^{2 n}-1$
Substituting $n = 1, 2, 3,….$
$P(1)=2^{2(1)}-1=4-1=3$
This is divisible by $3.$
$P(2)=2^{2(2)}-1=16-1=15$
This is divisible by $3.$
$P(3)=2^{2(3)}-1=256-1=255$
This is also divisible by $3.$
Assume that $P(n)$ is true for some natural number $k,$
i.e., $P(k): 2^{2k}- 1$ is divisible by $3,$
i.e., $2^{2k}- 1 = 3q$,
where $q \in N$
Now,
$P(k+1): 2^{2(k+1)}-1$
$=2^{2 k+2}-1$
$=2^{2 k} \times 2^2-1$
$=2^{2 k} \times 4-1$
$=3.2^{2 k}+\left(2^{2 k}-1\right)$
$=3.2^{2 k}+3 q$
$=3\left(2^{2 k}+q\right)=3m,$
where $m \in N$
Thus $P(k + 1)$ is true, whenever $P(k)$ is true.
Therefore, for any natural number $n, 2^{2n-1}$ is divisible by $3$
View full question & answer→MCQ 1561 Mark
$2^{3n}- 7n - 1$ is divisible by:
View full question & answer→MCQ 1571 Mark
Statement$-l:$ For every natural number $\text{n}\geq2,\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{n}}}>\sqrt{\text{n}}$
Statement$-2:$ For every natural number $\text{n}\geq2,\sqrt{\text{n}(\text{n}+1)}>\text{n}+1.$
- A
Statement$-1$ is true, Statement$-2$ is true; Statement$-2$ is a correct explanation for Statement$-1.$
- B
Statement$-1$ is true, Statement$-2$ is false.
- C
Statement$-1$ is false, Statement$-2$ is true.
- ✓
Statement$-1$ is true, Statement$-2$ is true; Statement$-2$ is not a correct explanation for Statement$-1$.
AnswerCorrect option: D. Statement$-1$ is true, Statement$-2$ is true; Statement$-2$ is not a correct explanation for Statement$-1$.
$\text{P}\ (\text{n})\ =\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{n}}}$
$\text{P}\ (2)\ =\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2}$
Let us assume that $P(k)$
$=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{k}}}>\sqrt{\text{k}}$ is true
$\therefore\ \text{P (k + 1)}=\ \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{\text{k}}}+\frac{1}{\sqrt{\text{k + 1}}}>\sqrt{\text{k+1}}$ has to be true
$\text{L.H.S}>\sqrt{\text{k}}+\frac{1}{\sqrt{\text{k + 1}}}=\frac{\sqrt{\text{k}\ (\text{k + 1)}+1}}{\sqrt{\ \text{k + 1}}}$
since $\sqrt{\text{k}\ (\text{k} + 1)}>\text{k}\ (\forall\text{k}\geq0)$
$\therefore\frac{\sqrt{\text{k}\ (\text{k + 1)}+1}}{\sqrt{\ \text{k + 1}}}>\frac{\text{k + 1}}{\sqrt{\text{k + 1}}}=\sqrt{\text{k + 1}}$
Let $\text{P}\ (\text{n})=\sqrt{\text{n}\ (\text{n + 1})}<(\text{k + 1})$
Statement$-1$ is correct.
$\text{P}\ (2)=\sqrt{2\times3}<3$
If $\text{P}\ \text{(k)}=\sqrt{\text{k (k + 1)}}<\text{(k + 1)}$ is true
Now $\text{P}\ \text{(k + 1)}=\sqrt{\text{(k + 1)}(\text{k + 2})}<\text{(k + 2)}$ has to be true
since $\text{(k + 1)}<\text{k + 2}$
$\therefore\sqrt{\text{(k + 1)}\text{(k + 2)}}<\text{(k + 2)}$
Hence Statement$-2$ is not correct explanation of Statement$-1.$
View full question & answer→MCQ 1581 Mark
For all $n \in N, 3n^5+ 5n^3+ 7n$ is divisible by:
AnswerGiven number $=3 n^5+5 n^2+7 n $
Let $ n=1,2,3,4,...... $
$ 3 n^5+5 n^3+7 n$
$=3 \times 1^2+5 \times 1^3+7 \times 1$
$=3+5+7$
$=15 $
$ 3 n^5+5 n^3+7 n$
$=3 \times 2^5+5 \times 2^3+7 \times 2$
$=3 \times 32+5 \times 8+7 \times 2$
$=96+40+14$
$=15 \times 10$
$=150$
$ 3 n^5+5 n^3+7 n$
$=3 \times 3^5+5 \times 3^3+7 \times 3$
$=3 \times 243+5 \times 27+7 \times 3$
$=729+135+21$
$=15 \times 59$
$=885$
Since, all these numbers are divisible by $15$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $15.$
View full question & answer→MCQ 1591 Mark
$2^{3n}- 7n - 1$ is divisible by:
View full question & answer→MCQ 1601 Mark
The remainder when $5^{99}$ is divided by $13,$ is:
View full question & answer→MCQ 1611 Mark
For each $n N \in , 3^{2n}- 1$ is divisible by:
View full question & answer→MCQ 1621 Mark
If $p(n): 49^\text{n}+16^{\text{n}}\lambda$ is divisible by $64$ for $\text{n}\in\text{N}$ is true, then the least negative integral value of $\lambda$ is:
Answer$ (49)^n+16 n-1 $
$ \Rightarrow(1+48)^n+16 n-1 $
$ \Rightarrow 1+48 n+\ldots 48^n+16 n-1 $
$ \Rightarrow 64 n+n C_2(48)^2+n C_3(48)^3+\ldots+(48)^n $
$ \Rightarrow 64\left(n+n C_2(6)^2+n C_3(6)^3 48+\ldots+(6)^n 8^{n-2}\right)$
$\therefore 49^n+ 16n - 1$ is divisible by $64$
View full question & answer→MCQ 1631 Mark
$7^{2n}+ 3^{n-1}⋅ 2^{3n-3}$ is divisible by:
AnswerLet $P(1) =7^{2n}+ 3^{n-1}⋅ 2^{3n-3}$
$P(1) = 50 \Rightarrow$ Divisible by $25$
View full question & answer→MCQ 1641 Mark
The sum of the series $1^2 + 2^2 + 3^2 + ……….. n^2$ is:
- A
$\text{n}(\text{n+1}(2\text{n+1)}$
- B
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{2}$
- C
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{3}$
- ✓
$\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
AnswerCorrect option: D. $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
Given, series is $1^2 + 2^2 + 3^2 + ……….. n^2$
$\text{sum=}\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}$
View full question & answer→MCQ 1651 Mark
Let $f(n) = 8n - 3n,$ if $n$ is odd natural number then $f(n)$ is divisible by:
Answer$f(n) = 8^n- 3^n$
Since, $n$ is odd
for $n = 1,$ we get $f(1) = 8^1− 3^1= 5$
for $n = 3,$ we get $f(3) = 8^3- 3^3= 5(97)$
for $n = 5,$ we get $f(3) = 8^5- 3^5= 5(6505)$
Therefore, by induction we can say that $f(n)$ is divisible by $5$ for odd $n.$
View full question & answer→MCQ 1661 Mark
In the following question, assuming the given statements to be true, find which of the conclusion among the given conclusions is/are definitely true and then give your answer accordingly.
Statement: $A ≥ P > T; V < B ≥ X; P = S; B = T$
Conclusion: $I. A > XII. P < B:$
AnswerCorrect option: D. Only $I$ is True.
Given statement: $A ≥ P > T; V < B ≥ X; P = S; B = T$
On combining: $A ≥ P = S > T = B ≥ X; V < B$
Conclusions:
$I. A > X \rightarrow$ True $(A \geq P = S > T = B \geq X)$
$II. P < B \rightarrow$ False $(P > T = B)$
Hence, only $I$ is True.
View full question & answer→MCQ 1671 Mark
The value of $(1+3/1)(1+5/4)(1+7/9)…(1+2n+1/n^2)$ is:
- A
$(n + 1)$
- B
$(n + 1)^2$
- C
$42(n+1)^2$
- ✓
View full question & answer→MCQ 1681 Mark
If $P(n)$ is statement such that $P(3)$ is true. assuming $P(k)$ is true $\Rightarrow P(k + 1)$ is true for all $k \geq 3,$ then $P(n)$ is true:
- A
for all $n$
- ✓
for $n ≥ 3$
- C
for $n ≥ 4$
- D
AnswerCorrect option: B. for $n ≥ 3$
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 $n$
In the given question $P(n)$ is true for $n = 3$
Assuming $P(k)$ is true
$\Rightarrow P(k + 1)$ is true for $k \geq 3$
Hence, By the Principle of Mathematical induction $P(n)$ is true for all $n \geq 3.$
View full question & answer→MCQ 1691 Mark
$S_n$ is divisible by the multiple of:
View full question & answer→MCQ 1701 Mark
$1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)}:$
AnswerCorrect option: A. $\{n(n + 3)\}/\{4(n + 1)(n + 2)\}$
Let $P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/\{n(n + 1)(n + 2)\} = \{n(n + 3)\}/\{4(n + 1)(n + 2)\}$
Putting $n = 1$ in the given statement, we get
$\text{LHS} = 1/(1 ∙ 2 ∙ 3) = 1/6$ and $\text{RHS} = \{1 \times (1 + 3)\}/[4 \times (1 + 1)(1 + 2)] = ( 1 \times 4)/(4 \times 2 \times 3) = 1/6.$
Therefore $\text{LHS = RHS.}$
Thus, the given statement is true for $n = 1,$
i.e., $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + … + 1/\{k(k + 1) (k + 2)\} = \{k(k + 3)\}/\{4(k + 1) (k + 2)\}.…(i)$
Now, $1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/\{k(k + 1) (k + 2)\} + 1/\{(k + 1) (k + 2) (k + 3)\}$
$= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}$
$= [\{k(k + 3)\}/\{4(k + 1)(k + 2)\} + 1/\{(k + 1)(k + 2)(k + 3)\}] [$using $(i)]$
$= {k(k + 3)^2 + 4}/\{4(k + 1)(k + 2) (k + 3)\}$
$= (k^3 + 6k^2 + 9k + 4)/\{4(k + 1) (k + 2) (k + 3)\}$
$= {(k + 1) (k + 1) (k + 4)}/\{4 (k + 1) (k + 2) (k + 3)\}$
$= \{(k + 1) (k + 4)\}/\{4(k + 2) (k + 3)\}$
$\Rightarrow P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/\{(k + 1) (k + 2) (k + 3)\}$
$= \{(k + 1) (k + 2)\}/\{4(k + 2) (k + 3)\}$
$\Rightarrow P(k + 1)$ 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 $n \in N.$
View full question & answer→MCQ 1711 Mark
If $x^{n-1}$ is divisible by $x - k,$ then the least positive integral value of $k$ is:
View full question & answer→MCQ 1721 Mark
$(1^2 + 2^2 + …… + n^2) ...........$ for all values of $n \in N:$
- A
$= n^3/3$
- B
$< n^3 /3$
- ✓
$> n^3/3$
- D
AnswerCorrect option: C. $> n^3/3$
Let $P(n): (1^2 + 2^2 + ….. + n^2) > n^3/3.$
When $= 1, \text{LHS} = 1^2 = 1$ and $\text{RHS} = 1^3/3 = 1/3.$
Since $1 > 1/3,$ it follows that $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): (1^2 + 2^2+ ….. + k^2 ) > k^3/3 …. (i)$
Now,
$1^2 + 2^2 + ….. + k^2+ (k + 1)^2$
$= \{1^2 + 2^2 + ….. + k^2 + (k + 1)^2$
$> k^3/3 + (k + 1)^3 [$using $(i)]$
$= 1/3 ∙ (k^3 + 3 + (k + 1)^2) = 1/3 ∙ \{k^2 + 3k^2 + 6k + 3\}$
$= 1/3[k^3 + 1 + 3k(k + 1) + (3k + 2)]$
$= 1/3 ∙ [(k + 1)^3 + (3k + 2)]$
$> 1/3(k + 1)^3P(k + 1):$
$1^2 + 2^2 + ….. + k^2 + (k + 1)^2$
$> 1/3 ∙ (k + 1)^3$
$P(k + 1)$ 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 $n \in N.$
View full question & answer→MCQ 1731 Mark
For every positive integer $n, 7^n– 3^n$ is divisible by:
AnswerLet $P(n) = 7^n- 3^n$
Substituting $n = 1, 2, 3,…$
$P(1) = 7^1– 3^1= 7 - 3 = 4 $
$P(2) = 7^2- 3^2= 49 - 9 = 40 $
$P(3) = 7^3-3^3= 343 - 27 = 316 $
Thus, for every positive integer $n, 7^n-3^n $ is divisible by $4.$
View full question & answer→MCQ 1741 Mark
For all $n \in N, 3 \times 5^{2n+1}+ 2^{3n+1}$ is divisible by:
View full question & answer→MCQ 1751 Mark
What is the sum of $1 + 2 + 3 + ... n\ ?$
AnswerCorrect option: C. $\frac{\text{n}(\text{n+1)}}{2}$
View full question & answer→MCQ 1761 Mark
If $49^n+ 16^n+ k$ is divisible by $64$ for $n \in N,$ then the least negative integral value of $k$ is:
View full question & answer→MCQ 1771 Mark
For all $n \geq 2,n \ n^2 (n^4 > 1)$ is divisible by:
View full question & answer→MCQ 1781 Mark
If $n(n^2 − 1)$ is divisible by $24,$ then which of the following statements is true?
- ✓
$n$ 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.
AnswerCorrect option: A. $n$ can be any odd integral value.
$n(n^2 − 1) = n ∗ (n − 1) ∗ (n + 1)$
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.
View full question & answer→MCQ 1791 Mark
If $P(n) = 2 + 4 + 6 +…+ 2n, n \in N ,P(k) = k(k + 1) + 2 \Rightarrow P(k + 1) = (k + 1)(k + 2) + 2$ for all $k \in N$. So, we can conclude that $P(n) = n (n + 1) 2$ for:
- A
all $n \in N$
- B
$n > 1$
- C
$n > 2$
- ✓
View full question & answer→MCQ 1801 Mark
For any natural number $n, 7^n– 2^n$ is divisible by
AnswerGiven, $7^n– 2^n$
Let $n = 1$
$7^n-2^n$
$=7^1-2^1$
$=7-2$
$=5$
which is divisible by $5$
Let $n = 2$
$7^n-2^n$
$=7^2-2^2$
$=49-4$
$=45$
which is divisible by $5$
Let $n = 3$
$7^n-2^n$
$=7^3-2^3$
$=343-8$
$=335$
which is divisible by $5$
Hence, for any natural number $n, 7^n-2^n$ is divisible by $5$
View full question & answer→MCQ 1811 Mark
For all positive integers $n > 1, \{x(x^{n-1}- na^{n-1} \})$ $a^n(n-1)$ is divisible by:
- A
$(x - a)^2$
- ✓
$x - a$
- C
$2(x - a)$
- D
$x + a$
AnswerCorrect option: B. $x - a$
View full question & answer→MCQ 1821 Mark
If $P(n) = 2 + 4 + 6 +…. + 2 n, n \in N,$ then $P(k) = k(k + 1) + 2 \Rightarrow P(k + 1) = (k + 1) (k + 2) + 2$ for all $k \in N.$ So, we can conclude that $P(n) = n(n + 1) + 2$ for:
- A
all $n \in N$
- B
$n > 1$
- C
$n > 2$
- ✓
View full question & answer→MCQ 1831 Mark
If $x^n- 1$ is divisible by $x - k,$ then the least positive integral value of $k$ is:
AnswerGiven,
$P(n): x^n- 1$ is divisible by $x - k$
Let us substitute $n = 1, 2, 3,..$
$ \Rightarrow P(1): x-1 $
$ \Rightarrow P(2): x^2-1=(x-1)(x+1) $
$ \Rightarrow P(3): x^3-1=(x-1)\left(x^2+x+1\right) $
$ \Rightarrow P(4): x^4-1=\left(x^2-1\right)\left(x^2+1\right)=(x-1)(x+1)\left(x^2+1\right) $
Therefore, the least positive integral value of $k$ is $1.$
View full question & answer→MCQ 1841 Mark
Let $P(n) : n^2 + n + 1$ is an even integer. If $P(k)$ is assumed true then $P(k + 1)$ is true. Therefore $P(n)$ is true.
- A
for $n > 1$
- B
for all $n Î N$
- C
for $n > 2$
- ✓
View full question & answer→MCQ 1851 Mark
For all $n \in N, 41^n– 14^n$ is a multiple of:
View full question & answer→