Mathematical Induction MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {\text{Let, }}m = 2k + 1,n = 2k - 1\left( {k \in N} \right) \cr & \therefore {m^2} - {n^2} = 4{k^2} + 1 + 4k - 4{k^2} + 4k - 1 = 8k \cr} $$
Hence, all the numbers of the form $$m^2 - n^2$$  are always divisible by 8.

$$\eqalign{ & {\text{For }}n = 1;{5^4} = 625 = \left( {624 + 1} \right) \cr & = \left( {48 \times 13} \right) + 1, \cr} $$
i.e., $$5^4$$ leaves 1 as remainder when divided by 13.

$$\eqalign{ & {\text{Let, }}P\left( n \right):{\left( {1 + x} \right)^n} \geqslant \left( {1 + nx} \right) \cr & {\text{For }}n = 1,{\left( {1 + x} \right)^1} = 1 + x \cr & = \,1 + 1 \cdot x \geqslant 1 + 1 \cdot x{\left( {1 + x} \right)^1} \geqslant 1 + 1 \cdot x \cr & {\text{For }}n = k,\,{\text{let }}P\left( k \right):{\left( {1 + x} \right)^k} \geqslant \left( {1 + kx} \right){\text{ is true}}{\text{.}} \cr & {\text{For }}n = k + 1,P\left( {k + 1} \right):{\left( {1 + x} \right)^{k + 1}} \geqslant \left\{ {1 + \left( {k + 1} \right)x} \right\} \cr} $$
is also true.
We will show $$P\left( {k + 1} \right)$$   is true.
Consider, $${\left( {1 + x} \right)^{k + 1}} = {\left( {1 + x} \right)^k} \cdot \left( {1 + x} \right) \geqslant \left( {1 + kx} \right)\left( {1 + x} \right)\left[ {{\text{If }}x > - 1} \right]$$
$$\eqalign{ & = 1 + x + kx + k{x^2} \geqslant 1 + x + kx\left[ {\because k > 0{\text{ and }}x > - 1} \right] \cr & = 1 + \left( {k + 1} \right)x \cr & {\text{Thus, }}{\left( {1 + x} \right)^{k + 1}} \geqslant 1 + \left( {k + 1} \right)x,{\text{ if }}x > - 1 \cr} $$

Let $$P\left( n \right):\frac{{{4^n}}}{{n + 1}} < \frac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}$$
For $$n = 2,P\left( 2 \right):\frac{{{4^2}}}{{2 + 1}} < \frac{{4!}}{{{{\left( 2 \right)}^2}}}$$
$$ \Rightarrow \frac{{16}}{3} < \frac{{24}}{4}$$
which is true.
Let for $$n = m \geqslant 2,P\left( m \right)$$    is true. i.e., $$\frac{{{4^m}}}{{m + 1}} < \frac{{\left( {2m} \right)!}}{{{{\left( {m!} \right)}^2}}}$$
Now, $$\frac{{{4^{m + 1}}}}{{m + 2}}$$
$$\eqalign{ & = \frac{{{4^m}}}{{m + 1}} \cdot \frac{{4\left( {m + 1} \right)}}{{m + 2}} < \frac{{\left( {2m} \right)!}}{{{{\left( {m!} \right)}^2}}} \cdot \frac{{4\left( {m + 1} \right)}}{{\left( {m + 2} \right)}} \cr & = \frac{{\left( {2m} \right)!\left( {2m + 1} \right)\left( {2m + 2} \right)4\left( {m + 1} \right){{\left( {m + 1} \right)}^2}}}{{\left( {2m + 1} \right)\left( {2m + 2} \right){{\left( {m!} \right)}^2}{{\left( {m + 1} \right)}^2}\left( {m + 2} \right)}} \cr & = \frac{{\left[ {2\left( {m + 1} \right)} \right]!}}{{{{\left[ {\left( {m + 1} \right)!} \right]}^2}}} \cdot \frac{{2{{\left( {m + 1} \right)}^2}}}{{\left( {2m + 1} \right)\left( {m + 2} \right)}} < \frac{{\left[ {2\left( {m + 1} \right)} \right]!}}{{{{\left[ {\left( {m + 1} \right)!} \right]}^2}}} \cr} $$
Hence, for $$n \geqslant 2,P\left( n \right)$$   is true.

The product of $$r$$ consecutive integers is divisible by $$r!.$$ Thus $$n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)$$     is divisible by $$4! = 24.$$

$$\eqalign{ & {10^n} + 3\left( {{4^{n + 2}}} \right) + 5\,\,\,\,{\text{Taking }}n = 2; \cr & {10^2} + 3 \times {4^4} + 5 = 100 + 768 + 5 = 873 \cr} $$
Therefore, this is divisible by 9.

Let $$P\left( n \right)$$  be the statement given by
$$P\left( n \right):{3^{2n}}$$   when divided by 8, the remainder is 1.
or $$P\left( n \right):{3^{2n}} = 8\lambda + 1$$    for some $$\lambda \in N$$
For $$n = 1,P\left( 1 \right):{3^2} = \left( {8 \times 1} \right) + 1 = 8\lambda + 1,\,$$       where $$\lambda = 1$$
$$\eqalign{ & \therefore P\left( 1 \right){\text{ is true}}{\text{.}} \cr & {\text{Let, }}P\left( k \right){\text{ be true}}{\text{.}} \cr} $$
Then, $${{\text{3}}^{2k}} = 8\lambda + 1$$   for some $$\lambda \in N\,\,\,\,.....\left( {\text{i}} \right)$$
We shall now show that $$P\left( {k + 1} \right)$$  is true, for which we have to show that $${{\text{3}}^{2\left( {k + 1} \right)}}$$  when divided by 8, the remainder is 1.
$$\eqalign{ & {\text{Now, }}{{\text{3}}^{2\left( {k + 1} \right)}} = {{\text{3}}^{2k}} \cdot {{\text{3}}^2} \cr & = \left( {8\lambda + 1} \right) \times 9\,\,\,\left[ {{\text{Using }}\left( {\text{i}} \right)} \right] \cr & = 72\lambda + 9 = 72\lambda + 8 + 1 = 8\left( {9\lambda + 1} \right) + 1 \cr & = 8\mu + 1,{\text{ where }}\mu = 9\lambda + 1 \in N \cr & \Rightarrow P\left( {k + 1} \right){\text{ is true}}{\text{.}} \cr} $$
Thus, $$P\left( {k + 1} \right)$$   is true, whenever $$P\left( {k} \right)$$  is true.
Hence, by the principle of mathematical induction $$P\left( {n} \right)$$  is true for all $$n \in N.$$

Check for $$n = 1, 2, 3, ..... ,$$    it is true for all $$n \in N.$$

Let, $$P\left( n \right):{7^n} - {3^n}$$   is divisible by 4.
For $$n = 1,$$
$$P\left( 1 \right):{7^1} - {3^1} = 4,$$    which is divisible by 4. Thus, $$P\left( n \right)$$  is true for $$n = 1.$$
Let $$P\left( k \right)$$  be true for some natural number $$k,$$ i.e., $$P\left( k \right):{7^k} - {3^k}$$   is divisible by 4.
We can write $${7^k} - {3^k} = 4d,$$   where $$d \in N\,\,\,\,.....\left( {\text{i}} \right)$$
Now, we wish to prove that $$P\left( {k + 1} \right)$$   is true
whenever $$P\left( k \right)$$  is true, i.e., $${7^{k + 1}} - {3^{k + 1}}$$   is divisible by 4.
$$\eqalign{ & {\text{Now, }}{7^{\left( {k + 1} \right)}} - {3^{\left( {k + 1} \right)}} \cr & = {7^{\left( {k + 1} \right)}} - 7 \cdot {3^k} + 7 \cdot {3^k} - {3^{\left( {k + 1} \right)}} \cr & = 7\left( {{7^k} - {3^k}} \right) + \left( {7 - 3} \right){3^k} = 7\left( {4d} \right) + 4 \cdot {3^k}\left[ {{\text{using }}\left( {\text{i}} \right)} \right] \cr & = 4\left( {7d + {3^k}} \right),{\text{ which is divisible by 4}}{\text{.}} \cr} $$
Thus, $$P\left( {k + 1} \right)$$  is true whenever $$P\left( {k} \right)$$  is true.
Therefore, by the principle of mathematical induction the statement is true for every positive integer $$n.$$

Putting $$n = 1$$  in $${11^{n + 2}} + {12^{2n + 1}}$$
We get, $${11^{1 + 2}} + {12^{2 \times 1 + 1}} = {11^3} + {12^3} = 3059,$$       which is divisible by 133.