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Algebra Level 4

If S n = ∑ r = 1 n t r = 2 n 3 + 9 n 2 + 13 n 6 S_n=\displaystyle\sum_{r=1}^{n} t_{r} = \dfrac{2n^3+9n^2+13n}{6} for all positive integers n n , evaluate ∑ r = 1 10 t r \displaystyle\sum_{r=1}^{10} \sqrt{t_r} .


The answer is 65.

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Parth Lohomi by Parth Lohomi , 20, India

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1 solution

Siddharth Bhatt
May 1, 2015

S n = n ( 2 n 2 + 9 n + 13 ) 6 S_n=\dfrac{n(2n^2+9n+13)}{6}

⟹ \implies t n = S n − S n − 1 = ( n + 1 ) 2 t_n=S_n-S_{n-1} = (n+1)^2

∑ r = 1 10 t r = ∑ n = 1 10 ( n + 1 ) = 65 \displaystyle\sum_{r=1}^{10}\sqrt{t_r}=\displaystyle\sum_{n=1}^{10} (n+1) =\boxed{65}

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Your first line is unnecessary.

Jake Lai - 5 years, 6 months ago

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