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A famous Hindu mathematician, Aryabhata, showed (in around 500 C.E.) you could add up the first n triangular numbers with a relatively simple formula: n(n + 1)(n + 2) =T + T2 + T3 + ... + Tn 6 Use this fact to find the sum of the first 53 triangular numbers. Answer: The sum of the first 53 square numbers is​

Respuesta :

mathstudent55

Answer:

26,235

Step-by-step explanation:

[tex] T_1 + T_2 + T_3 + ... + T_n = \dfrac{n(n + 1)(n + 2)}{6} [/tex]

[tex] T_1 + T_2 + T_3 + ... + T_{53} = \dfrac{53(53 + 1)(53 + 2)}{6} [/tex]

[tex] T_1 + T_2 + T_3 + ... + T_{53} = 26235 [/tex]

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