Question

F the sum of first `n' even natural numbers is equal to k times the sum of first `n' odd natural numbers, then k is equal to

Answer 1

The sum of first n even natural number is equal to $\frac{1}{n}+1$ the sum of first n odd natural numbers.

• Given that the sum of first n even natural numbers is k time the sum of first n odd natural numbers.
• First n even natural numbers is given by 2,4,6,8, and so on.
• Now the sum of first n even natural numbers is given by,

$S_n = \frac{n}{2}[2a + (n-1)d]$,

where a is the first term and d is the common difference and n is the number of terms in an AP.

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

• FIrst odd natural numbers is given by 1,3,5,7,9 and so on.
• Now the sum of first n odd natural numbers is given by,

$S_n = \frac{n}{2}[2a + (n-1)d]$

$S_n = \frac{n}{2}[2 + (n-1)2]$

$S_n = \frac{n}{2}[2n] = n^2$    (Equation 2)

• Now according to the given condition,

Sum of first n even natural numbers = k × Sum of first n odd natural numbers

• Substituting the values, we get

$n^2 + n = k (n^2)$

$n = k (n^2) - n^2$

$n = (k -1) n^2$

$1 = (k -1) n$

$\frac{1}{n} = k-1$

$\frac{1}{n}+1 = k$