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Write the sum using summation notation, assuming the suggested pattern continues. 16 + 25 + 36 + 49 + ... + n2 + ...
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answer choices: \[\sum_{n=5}^{\infty}n^2 \] \[\sum_{n=4}^{\infty}n+1^2\] \[\sum_{n=4}^{\infty}n-1^2\] \[\sum_{n=4}^{\infty}n^2\]
16 = 4^2 25 = 5^2 36= 6^2 49 = 7^2 You have a sum of the squares of integers starting with 4^2 The pattern is simply n^2, and the starting point is n = 4
so its the last one
yes
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