prove that the integral part of binomial expention is even 
                (5√5 + 11)2n+1

Joshi sir comment

Let x = (5√5 + 11)2n+1 and y = (5√5 - 11)2n+1

so x*y = 42n+1 = even number

similarly x - y = integer

if we consider the value of y , it will be (5*2.3-5)2n+1 = (0.5)2n+1  = fraction number less than 1   (approx)

so definitely y will be the fractional part of x, and difference of x and y is even so integer part of x will be even

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