It is given that the sequence $${x_{n}}$$ satisfies $$x_{1} = 0, x_{n+1} = x_{n} + 1 + 2\sqrt{1+ x_{n}}$$ for 𝑛 = 1, 2, ..... Then $$x_{31}$$ is ______________.

Correct Answer: 960

It is given that $$x_1=0$$

$$x_{n+1}=x_n+1+2\sqrt{1+x_n}$$

Thus, $$x_2=x_1+1+2\sqrt{1+x_1}$$ = 0+1+2 = 3

Similarly, $$x_3=x_2+1+2\sqrt{1+x_2}$$ = 3+1+4 = 8

Also, $$x_4=x_3+1+2\sqrt{1+x_3}$$ = 8+1+6 = 15

This forms a series: 0,3,8,15..

The general term of the series is $$T_n=n^2-1$$

Therefore, $$x_{31}$$ = 31*31-1 = 960

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