Sequence of roots

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Assume that x n = 3 + n + 3 n x_n=\sqrt 3+\sqrt n +\sqrt {3n} and S n = ( x n − 3 ) ( x n − n ) x n + 1 S_n=\frac {(x_n-\sqrt 3)(x_n-\sqrt n)} {x_n+1} , find the value of ∑ i = 1 100 S i 2 \sum_{i=1}^{100}S_{i^2}

5050 + 3 5050+\sqrt 3 100 3 100\sqrt 3 15150 15150 5050 3 5050\sqrt 3

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1 solution

Tom Engelsman
Feb 14, 2021

Let us take S i 2 = ( x i 2 − 3 ) ( x i 2 − i 2 ) x i 2 + 1 S_{i^2} = \frac{(x_{i^2}-\sqrt{3})(x_{i^2}-\sqrt{i^2})}{x_{i^2}+1} , where x i 2 = 3 + i 2 + 3 i 2 = 3 + ( 1 + 3 ) i x_{i^2} = \sqrt{3} + \sqrt{i^2} + \sqrt{3i^2} = \sqrt{3} + (1+\sqrt{3})i . This now yields:

S i 2 = [ ( 1 + 3 ) i + 3 ] 2 − ( i + 3 ) [ ( 1 + 3 ) i + 3 ] + 3 i 1 + [ ( 1 + 3 ) i + 3 ] = 3 + 3 3 + 1 ⋅ i = 3 + 3 3 + 1 ⋅ 3 − 1 3 − 1 ⋅ i = 3 − 3 + 2 3 3 − 1 ⋅ i = 3 i S_{i^2} = \frac{[(1+\sqrt{3})i + \sqrt{3}]^2 - (i+\sqrt{3})[(1+\sqrt{3})i+\sqrt{3}] + \sqrt{3} i}{1 + [(1+\sqrt{3})i + \sqrt{3}]} = \frac{\sqrt{3}+3}{\sqrt{3}+1} \cdot i = \frac{\sqrt{3}+3}{\sqrt{3}+1} \cdot \frac{\sqrt{3}-1}{\sqrt{3}-1} \cdot i = \frac{3-3 + 2\sqrt{3}}{3-1} \cdot i = \sqrt{3}i

and finally the summation:

Σ i = 1 100 3 i = 3 ⋅ ( 100 ) ( 101 ) 2 = 5050 3 . \Sigma_{i=1}^{100} \sqrt{3} i = \sqrt{3} \cdot \frac{(100)(101)}{2} = \boxed{5050\sqrt{3}}.

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why isnt the answer matching

Baibhab Chakraborty - 3 months, 1 week ago

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Seems like the admin had modified the question, it is not the actual answer too.

hello its me - 3 months ago

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