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CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 6 - MCQExams.com
CBSE
Class 11 Engineering Maths
Principle Of Mathematical Induction
Quiz 6
If $$n$$ is an odd positive integer, then $$a^n + b^n$$ is divisible by
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$$a - b$$
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$$a + b$$
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$$a^2 + b^2$$
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none of these
Explanation
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 $$n$$
The value of $$\displaystyle \frac{1^2}{1.3} + \frac{2^2}{3 . 5}+\dots+ \frac{n^2}{(2n - 1)(2 n + 1)}$$ is
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$$\displaystyle \frac{n (n + 1)}{2 (2n + 1)}$$
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$$\displaystyle \frac{n (n - 1)}{2 (2n - 1)}$$
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$$\displaystyle \frac{n^2 (n - 1)^2}{2 (2n + 1)}$$
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none of these
Explanation
$$\dfrac{n^{2}}{(2n-1)(2n+1)}$$
$$=\dfrac{4n^{2}}{4(2n-1)(2n+1)}$$
$$=\dfrac{1}{4}[\dfrac{4n^{2}-1+1}{4n^{2}-1}]$$
$$=\dfrac{1}{4}[1+\dfrac{1}{4n^{2}-1}]$$
$$=\dfrac{1}{4}[1+\dfrac{1}{2}(\dfrac{2n+1-(2n-1)}{4n^{2}-1})]$$
$$=\dfrac{1}{4}[1+\dfrac{1}{2}(\dfrac{1}{2n-1}-\dfrac{1}{2n+1})]$$
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
$$\dfrac{n}{4}+\dfrac{1}{8}\sum (\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$$
$$=\dfrac{n}{4}+\dfrac{2n}{8(2n+1)}$$
$$=\dfrac{n}{4}+\dfrac{n}{4(2n+1)}$$
$$=\dfrac{n(2n+1)+n}{4(2n+1)}$$
$$=\dfrac{n(2n+2)}{4(2n+1)}$$
$$=\dfrac{n(n+1)}{2(2n+1)}$$
The value of $$\displaystyle \frac{1}{1.2.3} + \frac{1}{2.3.4}+ ....+ \frac{1}{n (n + 1) (n + 2)}$$ is
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$$\displaystyle \frac{n(n + 3)}{4(n + 1) (n + 2)}$$
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$$\displaystyle \frac{n}{(n + 1)(n + 2)}$$
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$$\displaystyle \frac{n (n + 2)}{ (n + 1)(n + 3)}$$
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All of these
Explanation
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.
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
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$$\displaystyle \tan^{-1} \left ( \frac{n}{n + 2} \right )$$
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$$\displaystyle \tan^{-1} \left ( \frac{n+1}{n - 1} \right )$$
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$$\displaystyle \tan^{-1} \left ( \frac{n-1}{n + 2} \right )$$
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$$\displaystyle \tan^{-1} \left ( \frac{n+2}{n - 2} \right )$$
Explanation
$$T_n = \tan^{-1}\left(\cfrac{1}{1+n(1+n)}\right)= \tan^{-1}\left(\cfrac{(n+1)-n}{1+n(1+n)}\right)=\tan^{-1}(n+1)-\tan^{-1}n$$
Hence required sum is $$=\sum T_n$$
$$=[\tan^{-1}(2)-\tan^{-1}1]+[\tan^{-1}(3)-\tan^{-1}2]+........+\tan^{-1}(n)-\tan^{-1}(n-1)]+[\tan^{-1}(n+1)-\tan^{-1}n]$$
$$=\tan^{-1}(n+1)-\tan^{-1}1 = \tan^{-1}\left(\cfrac{n+1-1}{1+(n+1).1}\right)=\tan^{-1}\left(\cfrac{n}{n+2}\right)$$
By mathematical induction $$p^{n+1} + (p+1)^{2n -1}$$ is divisible by
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$$p^2 + p + 1$$
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$$p^2 + 1$$
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$$p+1$$
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None of these
Explanation
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 $$n\in N$$
If $$a_1 = 1, a_{n+1} = \dfrac{1}{n+1}a_n,\forall n \geq 1$$, then $$a_n \ =$$
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$$\displaystyle \frac{1}{n !}$$
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$$\displaystyle \frac{1}{(n + 2)!}$$
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$$\displaystyle \frac{1}{(n + 1)!}$$
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none of these
Explanation
$$a_1=1, a_2 = \cfrac{1}{1+1}a_{1}=\cfrac{1}{2}(1)=\cfrac{1}{2!}, a_3 = \cfrac{1}{3}.\cfrac{1}{2!} = \cfrac{1}{3!}$$
Similarly $$a_n = \cfrac{1}{n}.\cfrac{1}{(n-1)!} = \cfrac{1}{n!}$$
$$3 + 13 + 29 + 51 + 79 + ...$$ to $$n$$ terms $$=$$
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$$2n^2 + 7n^3$$
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$$n^2 + 5n^3$$
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$$n^3 + 2n^2$$
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none of these
Explanation
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
Using the principle of mathematical induction, find $$tan \alpha + 2 tan 2 \alpha + 2^2 tan 2^2 \alpha + ....$$ to $$n$$ terms:
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$$tan \alpha - 2^n \cdot tan (2^n \cdot \alpha)$$
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$$cot \alpha - 2^n \cdot cot (2^n \cdot \alpha)$$
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$$sec \alpha - 2^n \cdot sec (2^n \cdot \alpha)$$
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None of these
Explanation
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.
If $$p$$ is a prime number, then $$n^p -n$$ is divisible by $$p$$ for all $$n$$, where
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$$n\in N$$.
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$$n$$ is odd natural number.
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$$n$$ is even natural number.
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$$n$$ is not a composite number.
Explanation
$$n^{p}-n$$ 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 $$n\in N$$
For each $$n\, \epsilon\, N$$, then $$3^{2n\, +\, 1}\, +\, 1$$ is divisible by -
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$$2$$
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$$3$$
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$$7$$
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None of these
Explanation
$${ 3 }^{ 2n+1 }+1$$
Let $$n=1$$.
Then, $${ 3 }^{ 2n+1 }+1$$
$$={ 3 }^{ \left( 2\times 1 \right) +1 }+1$$
$$={ 3 }^{ 2+1 }+1$$
$${ 3 }^{ 3 }+1=28$$ which is divisible by $$2.$$
Let $$n=2$$.
Then, $${ 3 }^{ 2n+1 }+1$$
$${ 3 }^{ 5 }+1$$
$$243+1=244$$ which is divisible by $$2.$$
For positive integer n, $$3^n < n!$$ when
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$$n\geq 6$$
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$$n > 7$$
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$$n\geq 7$$
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$$n \leq 7$$
Explanation
For positive integer $$n$$,
$${ 3 }^{ n }<n!$$
Let $$n=1$$
$${ 3 }^{ 1 }<1!$$
$$={ 3 }<1$$ which is not possible.
Let $$n=2$$
$${ 3 }^{ 2 }<2!$$
$$={ 9 }<2$$ which is not possible.
Let $$n=6$$
$${ 3 }^{6 }<6!$$
$$={729 }<720$$ which is not possible.
Let $$n=7$$
$${ 3 }^{7 }<7!$$
$$={2187 }<5040$$ which is possible.
$$\therefore $$ For positive integer n, $${ 3 }^{ n }<n!$$ when $$n\ge 7$$
$$\forall$$ $$n\in N$$, $$\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}}$$ is
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$$\sqrt{n}$$
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$$\leq \sqrt{n}$$
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$$\gt\sqrt{n}$$
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none of these
Explanation
Let $$P(n)=\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}}$$ ....(1)
Put $$n=2$$ in (1)
$$P(2)=\displaystyle 1 + \frac{1}{\sqrt{2}} > \sqrt{2}$$
Put $$n=3$$ in (1)
$$P(3)=\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} >\sqrt{3}$$
So, let us assume
$$P(n)=\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}} >\sqrt{n}$$
So, we will check for $$P(n+1)$$
$$P(n+1)= \displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}} +\frac{1}{\sqrt{n+1}} $$
$$P(n+1)> \sqrt{n}+\frac{1}{\sqrt{n+1}} =\dfrac{\sqrt{n(n+1)}+1}{\sqrt{n+1}}$$ ....(2)
Since, $$\sqrt{n(n+1)}>n$$
$$\Rightarrow \sqrt{n(n+1)}+1 >n+1$$
$$\Rightarrow \dfrac{\sqrt{n(n+1)}+1}{\sqrt{n+1}}>\dfrac{n+1}{\sqrt{n+1}}$$
$$\Rightarrow \dfrac{\sqrt{n(n+1)}+1}{\sqrt{n+1}}>\sqrt{n+1}$$ ....(3)
From (2) and (3), it follows that
$$\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}>\sqrt{n+1}$$
Hence, by mathematical induction method, we can say that
$$\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}}>\sqrt{n}$$ for $$ n\in N, n\ge 2$$
Let $$\displaystyle p(n)=x\left(x^{n-1}-n\cdot a^{n-1}+a^{n}(n-1)\right)\>$$ is divisible by $$\left ( x-a \right )^{2}$$ for
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$$n> 1$$
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$$n> 2$$
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$$\forall \;n\;\in \;N$$
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None of these
Explanation
$$p(1)=x$$
Hence it is not divisible by $$(x-a)^{2}$$.
$$p(2)=x(x-2a+a^{2})$$
$$=x^{2}-2ax+a^{2}x$$
$$=(x-a)^{2}-a^{2}+a^{2}(x)$$
$$=(x-a)^{2}+a^{2}(x-1)$$ ...(i)
Hence $$p(2)$$ is also not divisible by $$(x-a)^{2}$$.
Now $$\{1,2\}$$ falls in the set of first n natural numbers.
Hence the correct answer is
Option D.
If $$10^n + 3\cdot 4^{n +2} + \lambda$$ is exactly divisible by $$9$$ for all $$n\in N$$, then the least positive integral value of $$\lambda$$ is
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$$5$$
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$$3$$
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$$7$$
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$$1$$
Explanation
Let $$P(n)=10^n + 3\cdot 4^{n +2} + \lambda$$
Since $$P(n)$$ is divisible by $$9$$ for all $$n\in N$$
$$\Rightarrow P(1)$$ will be also divisible by 9
$$\Rightarrow P(1)=10+3(4^3)+\lambda=202+\lambda=22\times 9+(\lambda+4)$$
Thus least positive integral value of $$\lambda$$ for which $$P(1)$$ can be divisible by $$9$$ is $$5$$
Hence option 'A" is correct choice.
For all $$n \in N$$, $$10^n + 3.4^{n+2} + 5$$ is divisible by
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$$23$$
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$$3$$
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$$9$$
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$$207$$
Explanation
$$10^n + 3.4^{n+2} + 5$$
Put $$n=1$$
$$10+3(64)+5=207$$ which is divisible by $$3,9,23,207$$
We will check for $$n=2$$
$$100+768+5=873$$
which is divisible by $$3$$ and $$9$$ only.
Every number divisible by $$9$$ is also divisible by $$3$$.
Now, let $$P(n)$$ is true
i.e. $$10^n + 3.4^{n+2} + 5$$ is divisible by $$9$$
$$\Rightarrow 10^n + 3.4^{n+2} + 5=9\lambda$$ ....(1)
Now, we will check for $$P(n+1)$$
Consider, $$10^{n+1} + 3.4^{n+3} + 5$$
$$=10^n10+3.4^{n+2}4+5$$
$$=10(9\lambda -3.4^{n+2}-5)+3.4^{n+2}4+5$$ (by (1))
$$=90\lambda-27.4^{n+2}-45$$
$$=9(10\lambda-3.4^{n+2}-5)$$
which is divisible by 9.
Hence, by mathematical induction given expression is divisible by $$3,9$$.
If
$$^{4n}{C_{{\text{2n}}}}{:^{2n}}{{\text{C}}_{\text{n}}} = \{ 1.3,.5,...(4{\text{n}} - 1)\} :{\{ 1,3,5....(2{\text{n}} - 1)\} ^\lambda },then\;\lambda = $$
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1
0%
2
0%
3
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none of these
Explanation
$$^{4n}C_{2n}:^{2n}C{n}=\left\{1,3,5,….\left(4n-1\right)\right\};\left\{1,3,5,…. \left(2n-1\right)\right\}$$
We have $$\dfrac{2n!}{n!}=\dfrac{2n\left(2n-1\right) \left(2n-1\right)….3.2.1}{n!}$$
$$=\left[2.4.6….\left(2n-2\right) \left(2n\right)\right]\left[1.3.5….\left(2n-1\right)\right]/n!$$
Writing the odd terms and even terms we get
$$\dfrac{2n!}{n!} = 2^{n}\left[1.2.3….\left(n-1\right).n\right] \left[1.3.5….\left(2n-1\right)\right]/n!$$
$$=\dfrac{2^{n}n! \left[1.3.5….\left(2n-1\right).n\right]}{n!}= \left[1.3.5….\left(2n-1\right)\right]$$
Replacing $$n$$ by $$2n$$,
$$\dfrac{4n!}{2n!}=2^{2n}\left[1.3.5….\left(4n-1\right)\right]$$
Now we have $$^{4n}C_{2n}:^{2n}C_{n}=\dfrac{4n!}{2n!2n!}\times\dfrac{n!n!}{2n!}=\dfrac{4n!}{2n!}.\left(\dfrac{n!}{2n!}\right)^{2}$$
Now $$^{4n}C_{2n}!^{2n}C_{n}=2^{2n}\dfrac{\left[1.3.5….\left(4n-1\right)\right]}{\left[2^{n}\left(1.3.5….(2n-1)\right)\right]}$$
$$= \dfrac{\left[1.3.5….\left(4n-1\right)\right]}{\left[ 1.3.5….\left(2n-1\right)\right]^{2}} \Rightarrow \lambda=2$$
If $$2^{83}+k $$ is divisible by 127, then the smallest positive integral value of k is:
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63
0%
31
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15
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64
Using principle of mathematics induct or for all
$$n\ \in \ N:1+2+3+....+n < \dfrac {1}{8}(2n+1)^{2}$$
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True
0%
False
For all $$n \in N ,$$
$$1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + 4 ^ { 2 } + \cdots + n ^ { 2 } = \dfrac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$$
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True
0%
False
$$\displaystyle{\frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 3 }{ { 3 }^{ 2 } } +......+\frac { 1 }{ { n }^{ 2 } } \le \frac { 3n+1 }{ 2n+2 } }$$ for every natural number $$n$$
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0%
True
0%
False
For any natural number $$n,$$ $$x^n y^n$$ is divisible by $$x y,$$ where $$x$$ and $$y$$ are any
integers with $$x \neq y.$$
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0%
True
0%
False
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