Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 19

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Question 1
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The value of $$\displaystyle \frac{1^2}{1.3} + \frac{2^2}{3 . 5}+\dots+ \frac{n^2}{(2n - 1)(2 n + 1)}$$ is

A
$$\displaystyle \frac{n (n + 1)}{2 (2n + 1)}$$

B
$$\displaystyle \frac{n (n - 1)}{2 (2n - 1)}$$

C
$$\displaystyle \frac{n^2 (n - 1)^2}{2 (2n + 1)}$$

D
none of these

Solution

$$\dfrac{n^{2}}{(2n-1)(2n+1)}$$

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

$$=\dfrac{1}{4}[\dfrac{4n^{2}-1+1}{4n^{2}-1}]$$

$$=\dfrac{1}{4}[1+\dfrac{1}{4n^{2}-1}]$$

$$=\dfrac{1}{4}[1+\dfrac{1}{2}(\dfrac{2n+1-(2n-1)}{4n^{2}-1})]$$

$$=\dfrac{1}{4}[1+\dfrac{1}{2}(\dfrac{1}{2n-1}-\dfrac{1}{2n+1})]$$

Consider
$$\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$$

$$=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}...\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$$

$$=1-\dfrac{1}{2n+1}$$

$$=\dfrac{2n}{2n+1}$$

Hence applying summation gives us

$$\dfrac{n}{4}+\dfrac{1}{8}\sum (\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$$

$$=\dfrac{n}{4}+\dfrac{2n}{8(2n+1)}$$

$$=\dfrac{n}{4}+\dfrac{n}{4(2n+1)}$$

$$=\dfrac{n(2n+1)+n}{4(2n+1)}$$

$$=\dfrac{n(2n+2)}{4(2n+1)}$$

$$=\dfrac{n(n+1)}{2(2n+1)}$$
Question 2
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By mathamatical induction $$n(n^2 - 1)$$ is divisible by

A
$$19$$

B
$$23$$

C
$$24$$

D
$$29$$

Solution

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.
Question 3
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If $$n$$ is an odd positive integer, then $$a^n + b^n$$ is divisible by

A
$$a - b$$

B
$$a + b$$

C
$$a^2 + b^2$$

D
none of these

Solution

Let, $$P(n) =a^n+b^n$$

$$P(1) = a+b$$, which is divisible by $$a+b$$

Now let $$P(k) =a^k+b^k$$ is divisible by $$a+b$$, where $$k$$ is an odd integer.

$$\Rightarrow a^k+b^k = (a+b)f(a,b)......................... (1)$$

Now, $$P(k+2)= a^{k+2}+b^{k+2} =a^2[(a+b)f(a,b)-b^k]+b^{k+2}$$

$$\quad \quad =a^2f(a,b)(a+b)-a^2b^k+b^{k+2}$$ from $$(1)$$

$$\quad \quad =a^2f(a,b)(a+b)-b^k(a^2-b^2)$$

$$\quad \quad =(a+b)\left[a^2f(a,b)-b^k(a-b)\right]$$, which is divisible by $$(a+b)$$

Hence $$a^n+b^n$$ is divisible by $$(a+b)$$ for all odd positive integral $$n$$
Question 4
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If $$a_1 = 1, a_{n+1} = \dfrac{1}{n+1}a_n,\forall n \geq 1$$, then $$a_n \ =$$

A
$$\displaystyle \frac{1}{n !}$$

B
$$\displaystyle \frac{1}{(n + 2)!}$$

C
$$\displaystyle \frac{1}{(n + 1)!}$$

D
none of these

Solution

$$a_1=1, a_2 = \cfrac{1}{1+1}a_{1}=\cfrac{1}{2}(1)=\cfrac{1}{2!}, a_3 = \cfrac{1}{3}.\cfrac{1}{2!} = \cfrac{1}{3!}$$
Similarly $$a_n = \cfrac{1}{n}.\cfrac{1}{(n-1)!} = \cfrac{1}{n!}$$
Question 5
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$$3 + 13 + 29 + 51 + 79 + ...$$ to $$n$$ terms $$=$$

A
$$2n^2 + 7n^3$$

B
$$n^2 + 5n^3$$

C
$$n^3 + 2n^2$$

D
none of these

Solution

Let $$S_{n}=3+13+29+51+79+\cdots +n terms$$

$$\because S_{1}=3$$ and $$S_{2}=3+13=16$$

Similarly $$S_{3}=3+13+29=45$$

By options, from $$(c)$$ as we have $$S_{n}=n^{3}+2n^{2}$$

check by putting $$ n=1, S_{1}=1^{3}+2.1^{2}=3$$

$$S_{2}=2^{3}+2.2^{2}=16 $$ and $$S_{3}=3^{3}+2.3^{2}=45$$

Thus option $$(c)$$ is correct
Question 6
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The value of $$\displaystyle \frac{1}{1.2.3} + \frac{1}{2.3.4}+ ....+ \frac{1}{n (n + 1) (n + 2)}$$ is

A
$$\displaystyle \frac{n(n + 3)}{4(n + 1) (n + 2)}$$

B
$$\displaystyle \frac{n}{(n + 1)(n + 2)}$$

C
$$\displaystyle \frac{n (n + 2)}{ (n + 1)(n + 3)}$$

D
All of these

Solution

Let $$P(n) : \displaystyle \frac{1}{a .2. 3} + \frac{1}{2.3.4} + .....+ \frac{1}{n(n + 1) (n+2)} = \frac{n(n+3)}{4(n + 1) (n + 2)}$$ ...... (1)

Step I: For $$n = 1$$,

L.H.S. of $$(1) = \displaystyle \frac{1}{1.2.3} = \frac{1}{6}$$

and R.H.S of $$(1) = \displaystyle \frac{1. (1 + 3)}{4(1 +1)(1+2)} = \frac{1}{6}$$

Therefore P(1) is true.

Step II : Assume that P(k) is true, then
$$\displaystyle P(k) : \frac{1}{1.2.3} + \frac{1}{2.3.4} + ....+ \frac{1}{k(k + 1)(k + 2)} = \frac{k(k + 3)}{4(k + 1)(k+2)}$$

Step III : For $$n = k +1$$,
$$\displaystyle P(k + 1) : \frac{1}{1.2.3} + \frac{1}{2.3.4} + .... + \frac{1}{k(k + 1)(k+2)} + \frac{1}{(k + 1)(k+2)(k+3)}$$

$$= \displaystyle \frac{(k + 1)(k+4)}{4(k + 2) (k + 3)}$$

$$\therefore $$ L.H.S $$= \displaystyle \frac{1}{1.2.3} + \frac{1}{2.3.4} + ..... + \frac{1}{k(k +1) (k + 2)} + \frac{1}{(k +1)(k + 2)(k+3)}$$

$$= \displaystyle \frac{k(k+3)}{4(k + 1)(k+2)} + \frac{1}{(k + 1)(k+2)(k + 3)}$$ (By assumption step)

$$= \displaystyle \frac{k (k + 3)^2 + 4}{4(k + 1)(k+2)(k+3)}$$

$$=\displaystyle \frac{k^3 + 6k^2 + 9k + 4}{4(k + 1)(k +2)(k+3)}$$

$$=\displaystyle \frac{(k + 1)^2(k + 4)}{4(k + 1)(k+2)(k+3)}$$

$$=\displaystyle \frac{(k + 1)(k +4)}{4 (k + 2)(k + 3)}$$

$$= R. H. S$$

Hence P(k + 1) is true. Hence by the principle of mathematical induction P(n) is true for all n $$\in$$ N.
Question 7
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The value of $$\displaystyle \tan^{-1} \left ( \frac{1}{3} \right ) + \tan^{-1} \left ( \frac{1}{7} \right ) +\dots+ \tan^{-1} \left ( \frac{1}{n^2 + n + 1} \right )$$ is

A
$$\displaystyle \tan^{-1} \left ( \frac{n}{n + 2} \right )$$

B
$$\displaystyle \tan^{-1} \left ( \frac{n+1}{n - 1} \right )$$

C
$$\displaystyle \tan^{-1} \left ( \frac{n-1}{n + 2} \right )$$

D
$$\displaystyle \tan^{-1} \left ( \frac{n+2}{n - 2} \right )$$

Solution

$$T_n = \tan^{-1}\left(\cfrac{1}{1+n(1+n)}\right)= \tan^{-1}\left(\cfrac{(n+1)-n}{1+n(1+n)}\right)=\tan^{-1}(n+1)-\tan^{-1}n$$

Hence required sum is $$=\sum T_n$$

$$=[\tan^{-1}(2)-\tan^{-1}1]+[\tan^{-1}(3)-\tan^{-1}2]+........+\tan^{-1}(n)-\tan^{-1}(n-1)]+[\tan^{-1}(n+1)-\tan^{-1}n]$$

$$=\tan^{-1}(n+1)-\tan^{-1}1 = \tan^{-1}\left(\cfrac{n+1-1}{1+(n+1).1}\right)=\tan^{-1}\left(\cfrac{n}{n+2}\right)$$
Question 8
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By mathematical induction $$p^{n+1} + (p+1)^{2n -1}$$ is divisible by

A
$$p^2 + p + 1$$

B
$$p^2 + 1$$

C
$$p+1$$

D
None of these

Solution

Let $$f\left( n \right)={ p }^{ n+1 }+{ \left( p+1 \right) }^{ 2n-1 }$$
we have $$f\left( 1 \right) ={ p }^{ 2 }+p+1$$ which is divisible by $${p}^{2}+p+1$$
Now, assume that $$f(m)$$ is divisible by $${p}^{2}+p+1$$
$$\therefore{ p }^{ m+1 }+{ \left( p+1 \right) }^{ 2m-1 }=k\left( { p }^{ 2 }+p+1 \right) $$ ...(1)
$$f\left( m+1 \right) ={ p }^{ m+2 }+{ \left( p+1 \right) }^{ 2m+2-1 }$$
$$={ p }^{ m+2 }+{ \left( p+1 \right) }^{ 2m-1 }.{ \left( p+1 \right) }^{ 2 }$$
$$={ p }^{ m+2 }+\left[ k\left( { p }^{ 2 }+p+1 \right) -{ p }^{ m+1 } \right] { \left( p+1 \right) }^{ 2 }$$
$$={ p }^{ m+2 }-{ \left( p+1 \right) }^{ 2 }{ p }^{ m+1 }+k{ \left( p+1 \right) }^{ 2 }\left( { p }^{ 2 }+p+1 \right) $$
$$={ p }^{ m+1 }\left( p-{ p }^{ 2 }-2p-1 \right) +k{ \left( p+1 \right) }^{ 2 }\left( { p }^{ 2 }+p+1 \right) $$
$$=-\left( { p }^{ 2 }+p+1 \right) \left[- k{ \left( p+1 \right) }^{ 2 }+{ p }^{ m+1 } \right] $$
Hence, $$f(m+1)$$ is divisible by $${p}^{2}+p+1$$
Hence by mathematical induction $$f(n)$$ is divisible by $${p}^{2}+p+1$$ for all $$n\in N$$
Question 9
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Using the principle of mathematical induction, find $$tan \alpha + 2 tan 2 \alpha + 2^2 tan 2^2 \alpha + ....$$ to $$n$$ terms:

A
$$tan \alpha - 2^n \cdot tan (2^n \cdot \alpha)$$

B
$$cot \alpha - 2^n \cdot cot (2^n \cdot \alpha)$$

C
$$sec \alpha - 2^n \cdot sec (2^n \cdot \alpha)$$

D
None of these

Solution

Let P(n): $$tan \alpha + 2 tan 2\alpha + 2^2 tan 2^2 \alpha + .... + 2^{n - 1} tan (2^{n -1} \alpha) = cot \alpha - 2^n \cdot cot (2^n \cdot \alpha)$$ ..... (1)

Step I: For $$n = 1$$,

L.H.S. of $$(1) = tan \alpha$$

$$= cot \alpha - cot \alpha + tan \alpha = cot \alpha- (cot \alpha - tan \alpha)$$

$$= cot \alpha- \displaystyle \left ( cot \alpha - \frac{1}{cot \alpha} \right )$$

$$= cot \alpha - 2 \left ( \dfrac{cot^2 \alpha - 1}{2 cot \alpha} \right )$$

$$= cot \alpha - 2 \left ( \dfrac{cos^2 \alpha - \sin ^2 \alpha}{2 cos \alpha \sin \alpha} \right )$$

$$= cot \alpha - 2 cot 2 \alpha$$

$$= R. H.S.$$ of (1)

Therefore, $$P(1)$$ is true.

Step II: Assume it is true for $$ n = k$$, then

$$P(k) : tan \alpha + 2 tan 2\alpha + 2^2 tan 2^2\alpha + .... + 2^{k - 1} tan (2^{k - 1} \alpha)$$

$$= cot \alpha - 2^k cot (2^k \alpha)$$

Step III: For $$n = k + 1$$,

$$P(k +1) : tan \alpha + 2 tan 2 \alpha + 2^2 tan 2^2 \alpha + ... + 2^{k - 1} tan (2^{k - 1} \alpha) + 2^k tan (2^k \alpha) = cot \alpha - 2^{k + 1} cot (2^{k + 1} \alpha)$$

$$L.H.S. = tan \alpha + 2 tan 2\alpha + 2^2 tan 2^2 \alpha + .... + 2^{k - 1} tan (2^{k - 1} \alpha) + 2^k tan (2^k \alpha)$$

$$= cot \alpha - 2^k cot (2^k \alpha) + 2^k tan (2^k \alpha)$$ (By assumption step)

$$= cot \alpha - 2^k (cot (2^k \alpha) - tan (2^k \alpha))$$

$$= cot \alpha - 2^k \cdot 2 \displaystyle \left ( \frac{cot^2 (2^k \alpha) - 1}{2 cot (2^k \alpha)} \right )$$

$$= cot \alpha - 2^{k + 1} \cdot cot (2 \cdot 2^k \alpha)$$

$$= cot \alpha - 2^{k + 1} \cdot cot (2^{k + 1} \alpha)$$

$$= R.H.S.$$

This show that the result is true for $$n = k + 1$$.

Hence by the principle of mathematical induction, the result is true for all n $$\in$$ N.
Question 10
1 / -0

If $$p$$ is a prime number, then $$n^p -n$$ is divisible by $$p$$ for all $$n$$, where

A
$$n\in N$$.

B
$$n$$ is odd natural number.

C
$$n$$ is even natural number.

D
$$n$$ is not a composite number.

Solution

$$n^{p}-n$$ is divisible by $$p$$ for $$n=1$$
Let, $$k^{p}-k=p\lambda$$, where $$k\in N$$ and $$\lambda \in Z$$

Now, $$\left (k+1 \right )^{p}-\left (k+1 \right )=k^{p}+^{p}\textrm{C}_{1}k^{p-
1}+.........+^{p}\textrm{C}_{p}-1-k$$

$$\left (k^{p}-k \right )+\left (^{p}\textrm{C}_{1}k^{p-1}+...+^{p}\textrm{C}_{p-1}k
\right )$$

$$=p\lambda +\left (^{p}\textrm{C}_{1}k^{p-1}+...+^{p}\textrm{C}_{p-1}k \right )$$
which is divisible by $$p$$
Reason :$$^{p}\textrm{C}_f$$ is always divisible by by p when $$f\neq p,0$$ because p is a prime number as the denominator will not have any factors or multiples of p.

So by mathematical induction, $$n^{p}-n$$ is divisible by $$p$$ for all $$n\in N$$

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Principle of Mathematical Induction Test - 19

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Principle of Mathematical Induction Test - 19

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Question 1

$$\displaystyle \frac{n (n + 1)}{2 (2n + 1)}$$

$$\displaystyle \frac{n (n - 1)}{2 (2n - 1)}$$

$$\displaystyle \frac{n^2 (n - 1)^2}{2 (2n + 1)}$$

none of these

Solution

Question 2

$$19$$

$$23$$

$$24$$

$$29$$

Solution

Question 3

$$a - b$$

$$a + b$$

$$a^2 + b^2$$

none of these

Solution

Question 4

$$\displaystyle \frac{1}{n !}$$

$$\displaystyle \frac{1}{(n + 2)!}$$

$$\displaystyle \frac{1}{(n + 1)!}$$

none of these

Solution

Question 5

$$2n^2 + 7n^3$$

$$n^2 + 5n^3$$

$$n^3 + 2n^2$$

none of these

Solution

Question 6

$$\displaystyle \frac{n(n + 3)}{4(n + 1) (n + 2)}$$

$$\displaystyle \frac{n}{(n + 1)(n + 2)}$$

$$\displaystyle \frac{n (n + 2)}{ (n + 1)(n + 3)}$$

All of these

Solution

Question 7

$$\displaystyle \tan^{-1} \left ( \frac{n}{n + 2} \right )$$

$$\displaystyle \tan^{-1} \left ( \frac{n+1}{n - 1} \right )$$

$$\displaystyle \tan^{-1} \left ( \frac{n-1}{n + 2} \right )$$

$$\displaystyle \tan^{-1} \left ( \frac{n+2}{n - 2} \right )$$

Solution

Question 8

$$p^2 + p + 1$$

$$p^2 + 1$$

$$p+1$$

None of these

Solution

Question 9

$$tan \alpha - 2^n \cdot tan (2^n \cdot \alpha)$$

$$cot \alpha - 2^n \cdot cot (2^n \cdot \alpha)$$

$$sec \alpha - 2^n \cdot sec (2^n \cdot \alpha)$$

None of these

Solution

Question 10

$$n\in N$$.

$$n$$ is odd natural number.

$$n$$ is even natural number.

$$n$$ is not a composite number.

Solution

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