Sequences

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2305
Shaswata Roy ·

xk+1 = xk(xk+1)

1xk+1 = 1xk(xk+1) = 1xk - 1xk+1

or,1xk+1 = 1xk - 1xk+1

The required sum is =

1x1+1 + 1x2+1 + 1x3+1 + ..........+1x100+1
= 1x1-1x2+........+1x99-1x100
=1x1 - 1x100

x3 >1
If xk >1 then xk+1 = xk(xk+1)>1

Using induction we can say that x100 >1

Therefore required sum = 1x1 - 1x100 = 2 - 1x100.

Since x100 >1 , 1x100<1

Therefore the integer part of the sum is = 1.

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