If the sum of first n even natural number is equal to k times the sum of first n odd number then find k
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Answer:
Here's the answer : (1+n)/n
Step-by-step explanation:
first even natural number is : 2
first odd natural number is : 1
sum of n even natural numbers = n/2{2*2 +d(n-1)}
sum of n odd natural numbers = n/2{2*1 +d(n-1)}
ATP,
n/2{2*2 +d(n-1)} = k * n/2{2*1 +d(n-1)}
so, k = n/2{2*2 +d(n-1)} / n/2{2*1 +d(n-1)}
k = 4 + d(n-1) / 2 + d(n-1)
here, d = 2 (as, the difference between 2 consecutive odd or, even number is 2)
so, k = 4 + 2(n-1) / 2 + 2(n-1)
= 2 + n -1 / 1 + n -1
= (1 + n)/ n
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