Incongruity between summation and proof result

Question

A website made the statement $P(n) = \sum_{i=0}^n i^2 = \frac{1}{6}n(n+1)(2n+1)$.

If I want to know the value at the 3rd position in the sequence is, I just plug $3$ into the equation:

$$P(3) = \sum_{n=0}^3 = i^2 = \frac{1}{6}3(3+1)(2(3)+1)$$

And the result is $\frac{84}{6} = 14$. Why is it not 4, which is the third integer generated by the summation: $0, 1, 4, 9, 16...$?

Answer 1

You seem to be confusing sequences and series:

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A sequence is a list of numbers separated with commas. For example: $$ 1, 4, 9, 16, \ldots $$ Notice that the $n$th term in the sequence is given by $t_n = n^2$.

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A finite series is the sum of the first few terms in a sequence. For example: \begin{align*} s_1 &= 1 \\ s_2 &= 1 + 4 = 5 \\ s_3 &= 1 + 4 + 9 = 14 \\ s_4 &= 1 + 4 + 9 + 16 = 30 \\ \end{align*}

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You are being asked to prove that the sum of the first $n$ terms in the sequence is given by the formula: $$ s_n = \frac{1}{6}n(n+1)(2n+1) $$

Answer 2

The sequence you have written is correctly described by $\lbrace i^2\rbrace_{i \in \mathbb{N}}$, instead of $\lbrace \sum_0^i k^2\rbrace_{i \in \mathbb{N}}$. In fact, $0 + 1^2 + 2^2 + 3^2 = 1+4+9 = 14$ gives the 3rd term of the latter sequence, so it is correct.