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If mgt1 and ninN, such that 1^(m)+2^(m...

If `mgt1` and `ninN`, such that
`1^(m)+2^(m)+3^(m)+...+n^(m)gtn((n+1)/(k))^(m)` Then, k=

A

2

B

n

C

m

D

1

Text Solution

AI Generated Solution

The correct Answer is:
To solve the given inequality \( 1^m + 2^m + 3^m + \ldots + n^m > n \left( \frac{n+1}{k} \right)^m \), we will follow the steps outlined below. ### Step 1: Understanding the Arithmetic Mean Inequality We know from the Arithmetic Mean - Geometric Mean (AM-GM) inequality that: \[ \frac{1^m + 2^m + 3^m + \ldots + n^m}{n} \geq \left( \frac{1 + 2 + 3 + \ldots + n}{n} \right)^m \] This means that the arithmetic mean of the \( m \)-th powers of the first \( n \) natural numbers is greater than or equal to the \( m \)-th power of the arithmetic mean of the first \( n \) natural numbers. ### Step 2: Calculate the Arithmetic Mean of the First \( n \) Natural Numbers The sum of the first \( n \) natural numbers is given by: \[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \] Thus, the arithmetic mean is: \[ \frac{1 + 2 + 3 + \ldots + n}{n} = \frac{n(n + 1)/2}{n} = \frac{n + 1}{2} \] ### Step 3: Substitute into the AM-GM Inequality Substituting this back into our inequality gives: \[ 1^m + 2^m + 3^m + \ldots + n^m > n \left( \frac{n + 1}{2} \right)^m \] ### Step 4: Rearranging the Inequality We can rearrange our inequality to compare it with the original inequality: \[ 1^m + 2^m + 3^m + \ldots + n^m > n \left( \frac{n + 1}{k} \right)^m \] This implies: \[ n \left( \frac{n + 1}{2} \right)^m > n \left( \frac{n + 1}{k} \right)^m \] ### Step 5: Canceling \( n \) (Assuming \( n > 0 \)) Since \( n \) is a natural number (and thus greater than 0), we can safely divide both sides by \( n \): \[ \left( \frac{n + 1}{2} \right)^m > \left( \frac{n + 1}{k} \right)^m \] ### Step 6: Taking the \( m \)-th Root Taking the \( m \)-th root of both sides (since \( m > 1 \)): \[ \frac{n + 1}{2} > \frac{n + 1}{k} \] ### Step 7: Cross-Multiplying Cross-multiplying gives: \[ (n + 1)k > 2(n + 1) \] Assuming \( n + 1 > 0 \), we can divide both sides by \( n + 1 \): \[ k > 2 \] ### Step 8: Finding the Value of \( k \) Since we need to find the value of \( k \) such that the original inequality holds, we can conclude that: \[ k = 2 \] Thus, the value of \( k \) is \( 2 \). ### Final Answer \[ k = 2 \]

To solve the given inequality \( 1^m + 2^m + 3^m + \ldots + n^m > n \left( \frac{n+1}{k} \right)^m \), we will follow the steps outlined below. ### Step 1: Understanding the Arithmetic Mean Inequality We know from the Arithmetic Mean - Geometric Mean (AM-GM) inequality that: \[ \frac{1^m + 2^m + 3^m + \ldots + n^m}{n} \geq \left( \frac{1 + 2 + 3 + \ldots + n}{n} \right)^m \] This means that the arithmetic mean of the \( m \)-th powers of the first \( n \) natural numbers is greater than or equal to the \( m \)-th power of the arithmetic mean of the first \( n \) natural numbers. ...
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