Let P n -2n- n- - n- N and 2k- k- - n-k- then which of the following is true - k- 2-

The correct option is A 2^k+1> 2k> k+1 2 ^ k gt;k , ∀ n=k So, 2 ^ k+1 gt;k+1 #160; #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

Let P n :2n...

Question

Let $P (n) : 2^{n} > n, \forall n \in N$ and $2^{k} > k, \forall n = k$ , then which of the following is true $\forall k \geq 2$ ?

2^{k} > 5 k > 1

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2^{k + 1} > 2 k > k + 1

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2^{k} > 2 (k + 1) > k

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None of these.

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Solution

The correct option is A $2^{k + 1} > 2 k > k + 1$
$2^{k} > k$ , $\forall n = k$
So, $2^{k + 1} > k + 1$ ...(i)

Also, $k + k > k + 1$ , $\forall k \geq 2$
$\Rightarrow 2 k > k + 1$ ...(ii)

And $2^{k + 1} > 2^{k} > 2 k + 1$ ...(iii)
From (i), (ii) and (iii), we get option $B$ .

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Similar questions

Q. If

D_{k} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & n & n 2 k & n^{2} + n + 1 & n^{2} + n 2 k - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

and

\sum_{k = 1}^{n} D_{k} = 56

, then

n

equals ?

Q. If

S_{k} = sin (\frac{2 π k}{n + 1}) - i cos (\frac{2 π k}{n + 1})

and

P_{k} = cos (\frac{π}{2^{k}}) + i sin (\frac{π}{2^{k}}),

Let

m = ∣ ∣ ∣ n \sum k = 1 S_{k} + \infty \prod k = 1 P_{k} ∣ ∣ ∣

, then the value of

2 m^{2}

Q. If

S_{k} = sin (\frac{2 π k}{n + 1}) - i cos (\frac{2 π k}{n + 1})

and

P_{k} = cos (\frac{π}{2^{k}}) + i sin (\frac{π}{2^{k}}),

Let

m = ∣ ∣ ∣ n \sum k = 1 S_{k} + \infty \prod k = 1 P_{k} ∣ ∣ ∣

, then the value of

2 m^{2}

Q. The system of simultaneous equations

3 m + n = 1

and

(2 k - 1) m + (k - 1) n = 2 k + 1

, is inconsistent. What is the value of 'k'?

If $D_{k} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & n & n 2 k & n^{2} + n + 1 & n^{2} + n 2 k - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ and $n \sum k = 1 D_{k} = 56,$ where $n \in N,$ then $n$ is equal to

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Let P(n):2n>n,∀n∈N and 2k>k,∀n=k, then which of the following is true ∀k≥2?

The correct option is A 2k+1>2k>k+12k>k, ∀n=kSo, 2k+1>k+1 ...(i)Also, k+k>k+1, ∀k≥2⇒2k>k+1 ...(ii)And 2k+1>2k>2k+1 ...(iii)From (i), (ii) and (iii), we get option B.

Let $P (n) : 2^{n} > n, \forall n \in N$ and $2^{k} > k, \forall n = k$ , then which of the following is true $\forall k \geq 2$ ?

The correct option is A $2^{k + 1} > 2 k > k + 1$
$2^{k} > k$ , $\forall n = k$
So, $2^{k + 1} > k + 1$ ...(i)

Also, $k + k > k + 1$ , $\forall k \geq 2$
$\Rightarrow 2 k > k + 1$ ...(ii)

And $2^{k + 1} > 2^{k} > 2 k + 1$ ...(iii)
From (i), (ii) and (iii), we get option $B$ .