find the ratio of n even numbers to the n odd number
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Sum of n even number = 2+4+6+…n
a = 2, d = 2
Sn = (n/2)(2a+(n-1)d)
= (n/2)(4+(n-1)2)
= (n/2)(2+2n)
= n(1+n)
Sum of n odd number = 1+3+5+…2n-1
a = 1, d = 2
Sn = (n/2)(2a+(n-1)d)
= ((n)/2)(2+(n-1)2)
= n2
Ratio of the sums of first n even number and n odd number = n(1+n)/n2
= (n+1)/n
HOPE IT HELPS
THANK YOU
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