1.
If $$\frac{{\left( {n + 2} \right)!}}{{6\left( {n - 1} \right)!}}$$ divisible by $$n,n \in N\,$$ and $$1 \leqslant n \leqslant 9,$$ then $$n$$ is A.
4
B.
2
C.
6
D.
1
Answer :
1
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If $$n = 1,$$ then $$\frac{{\left( {n + 2} \right)!}}{{6\left( {n - 1} \right)!}} = \frac{{3!}}{{6 \times 0!}} = \frac{6}{6} = 1,$$ divisible by 1.
2.
If \[A = \left[ {\begin{array}{*{20}{c}}
1&0\\
1&1
\end{array}} \right]{\rm{and }}\,\,I = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right],\] then which one of the following holds for all $$n \geqslant 1,$$ by the principle of mathematical induction A.
$${A^n} = nA - \left( {n - 1} \right)I$$
B.
$${A^n} = {2^{n - 1}}A - \left( {n - 1} \right)I$$
C.
$${A^n} = nA + \left( {n - 1} \right)I$$
D.
$${A^n} = {2^{n - 1}}A + \left( {n - 1} \right)I$$
Answer :
$${A^n} = nA - \left( {n - 1} \right)I$$
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We observe that
3.
The inequality $$n! > 2^{n - 1}$$ is true for A.
$$n > 2$$
B.
$$n \in N$$
C.
$$n > 3$$
D.
None of these
Answer :
$$n > 2$$
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$$\eqalign{
& {\text{Let, }}P\left( n \right) \equiv n! > {2^{n - 1}};P\left( 3 \right) \equiv 6 > 4 \cr
& {\text{Let, }}P\left( k \right) \equiv k! > {2^{k - 1}}{\text{ is true}}{\text{.}} \cr
& \therefore P\left( {k + 1} \right) = \left( {k + 1} \right)! = \left( {k + 1} \right)k! > \left( {k + 1} \right){2^{k - 1}} > {2^k}\left( {{\text{as }}k + 1 > 2} \right) \cr} $$
4.
If $$P = n\left( {{n^2} - {1^2}} \right)\left( {{n^2} - {2^2}} \right)\left( {{n^2} - {3^2}} \right).....\left( {{n^2} - {r^2}} \right),n > r,n \in N$$ then $$P$$ is necessarily divisible by A.
$$\left( {2r + 2} \right)!$$
B.
$$\left( {2r + 4} \right)!$$
C.
$$\left( {2r + 1} \right)!$$
D.
None of these
Answer :
$$\left( {2r + 1} \right)!$$
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$$\eqalign{
& P = n\left( {n + 1} \right)\left( {n - 1} \right)\left( {n + 2} \right)\left( {n - 2} \right).....\left( {n + r} \right)\left( {n - r} \right) \cr
& = \left\{ {n\left( {n + 1} \right)\left( {n + 2} \right).....\left( {n + r} \right)} \right\}\left\{ {\left( {n - 1} \right)\left( {n - 2} \right).....\left( {n - r} \right)} \right\} \cr
& = \left( {n + r} \right)\left( {n + r - 1} \right).....\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right).....\left( {n - r} \right) \cr} $$
5.
$${10^n} + 3\left( {{4^{n + 2}}} \right) + 5$$ is divisible by $$\left( {n \in N} \right)$$ A.
7
B.
5
C.
9
D.
17
Answer :
9
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$$\eqalign{
& {10^n} + 3\left( {{4^{n + 2}}} \right) + 5 \cr
& {\text{Taking, }}n = 2;{10^2} + 3 \times {4^4} + 5 \cr
& = 100 + 768 + 5 = 873 \cr} $$
6.
A student was asked to prove a statement $$P\left( n \right)$$ by induction. He proved that $$P\left( {k + 1} \right)$$ is true whenever $$P\left( 5 \right)$$ is true for all $$k > 5 \in N$$ and also that $$P\left( 5 \right)$$ is true. On the basis of this he could conclude that $$P\left( n \right)$$ is true A.
for all $$n \in N$$
B.
for all $$n > 5$$
C.
for all $$n \geqslant 5$$
D.
for all $$n < 5$$
Answer :
for all $$n \geqslant 5$$
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Since, $$P\left( 5 \right)$$ is true and $$P\left( {k + 1} \right)$$ is true, whenever $$P\left( k \right)$$ is true.
7.
If $$P\left( n \right):$$ "$$46^n + 19^n + k$$ is divisible by 64 for $$n \in N$$ " is true, then the least negative integral value of $$k$$ is A.
$$ - 1$$
B.
$$1$$
C.
$$2$$
D.
$$ - 2$$
Answer :
$$ - 1$$
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For $$n = 1,P\left( 1 \right):65 + k$$ is divisible by 64.
8.
If $$\frac{1}{{2 \times 4}} + \frac{1}{{4 \times 6}} + \frac{1}{{6 \times 8}} + .....\,n{\text{ terms}} = \frac{{kn}}{{n + 1}},$$ then $$k$$ is equal to A.
$$\frac{1}{4}$$
B.
$$\frac{1}{2}$$
C.
$$1$$
D.
$$\frac{1}{8}$$
Answer :
$$\frac{1}{4}$$
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$$\eqalign{
& \frac{{kn}}{{n + 1}} = \left[ {\frac{1}{{2 \times 4}} + \frac{1}{{4 \times 6}} + \frac{1}{{6 \times 8}} + .....\,n{\text{ terms}}} \right] \cr
& = \frac{1}{2}\left[ {\frac{{4 - 2}}{{2 \times 4}} + \frac{{6 - 4}}{{4 \times 6}} + \frac{{8 - 6}}{{6 \times 8}} + ..... + \frac{{2n + 2 - 2n}}{{2n\left( {2n + 2} \right)}}} \right] \cr
& = \frac{1}{2}\left[ {\frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{6} + \frac{1}{6} - \frac{1}{8} + ..... + \frac{1}{{2n}} - \frac{1}{{2n + 2}}} \right] \cr
& = \frac{1}{2}\left[ {\frac{1}{2} - \frac{1}{{2\left( {n + 1} \right)}}} \right] \cr
& = \frac{n}{{4\left( {n + 1} \right)}} \cr
& \Rightarrow k = \frac{1}{4} \cr} $$
9.
Using mathematical induction, the numbers $${a_n} 's$$ are defined $${a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n},\left( {n \geqslant 0} \right).$$ Then, $$a_n$$ is equal to A.
$${n^3} + {n^2} + 1$$
B.
$${n^3} - {n^2} + 1$$
C.
$${n^3} - {n^2} $$
D.
$${n^3} + {n^2}$$
Answer :
$${n^3} - {n^2} + 1$$
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$$\eqalign{
& {\text{Given, }}{a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n} \cr
& \Rightarrow {a_1} = 3\left( 0 \right) + 0 + {a_0} = 1 \cr
& \Rightarrow {a_2} = 3{\left( 1 \right)^2} + 1 + {a_1} = 3 + 1 + 1 = 5 \cr
& {\text{From option }}\left( B \right), \cr
& {\text{Let, }}P\left( n \right) = {n^3} - {n^2} + 1 \cr
& \therefore P\left( 0 \right) = 0 - 0 + 1 = 1 = {a_0} \cr
& P\left( 1 \right) = {1^3} - {1^2} + 1 = 1 = {a_1} \cr
& {\text{and }}P\left( 2 \right) = {\left( 2 \right)^3} - {\left( 2 \right)^2} + 1 = 5 = {a_2} \cr
& \therefore {a_n} = {n^3} - {n^2} + 1 \cr} $$
10.
If $$n$$ is a natural number, then A.
$${1^2} + {2^2} + ..... + {n^2} < \frac{{{n^3}}}{3}$$
B.
$${1^2} + {2^2} + ..... + {n^2} = \frac{{{n^3}}}{3}$$
C.
$${1^2} + {2^2} + ..... + {n^2} > {n^3}$$
D.
$${1^2} + {2^2} + ..... + {n^2} > \frac{{{n^3}}}{3}$$
Answer :
$${1^2} + {2^2} + ..... + {n^2} > \frac{{{n^3}}}{3}$$
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By taking option $$\left( d \right),$$