IPMAT Indore 2021 (SA) - It is given that the sequence x_n satisfies x_1 = 0, x_n+1 = x_n + 1 + 2√(1+x_n) for n = 1,2,... Then x_31 is _______ | PYQs + Solutions | AfterBoards
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IPMAT Indore 2021

Algebra
>
Progression & Series

Medium

It is given that the sequence {xnx_n} satisfies x1=0,xn+1=xn+1+2√(1+xn)x_1 = 0, x_{n+1} = x_n + 1 + 2√(1+x_n) for n=1,2,...n = 1,2,... Then x31x_{31} is _______

Entered answer:

Correct Answer: 960

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