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In his book, Calculus Vol. 1, Tom Apostol mentions that adding $k^2$ to the predicate

$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 < \frac{k^3}{3}$$

gives the inequality

$$ 1^2 + 2^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$

Why does the RHS $$1^2 + 2^2 + \cdots + (k - 1)^2 $$ become $$ 1^2 + 2^2 + \cdots + k^2 $$ and not $$ 1^2 + 2^2 + \cdots + (k - 1)^2 + k^2$$ when adding $k^2$?

Thank you.

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3 Answers 3

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The notation does not matter. $1^2 + 2^2 + \cdots + (k - 1)^2 + k^2$ and $1^2 + 2^2 + \cdots + k^2$ both stand for the same value (under the appropriate interpretation of $\cdots$). Both stand for $\displaystyle \sum_{j=1}^{k}j^2$.

Example:

$1^2+2^2+3^2 + \cdots + 7^2 = 1^2+2^2+3^2 + \cdots + 6^2 + 7^2 = 1^2+2^2+3^2 + 4^2+5^2 + 6^2 + 7^2 = \displaystyle \sum_{j=1}^{7}j^2=140$

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I will give you an example.

1+2+3+........+99+100 is same as 1+2+.....+100.

And, I believe this is quite obvious.

Hence, both of them have the same value whatever way you decide to solve it by.

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The reason why,$$1^2 + 2^2 + \cdots + (k-1)^2,$$ becomes, $$1^2 + 2^2 + \cdots + (k - 1)^2+ k^2,$$ is because you're adding the $k$th term squared on both sides. Thus, instead of having the sum up to the $k-1$ th term squared, now you have the sum up to the $k$th term squared.

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  • $\begingroup$ Thank you. So whenever you add a term to a sequence such as the one I mentioned, that means that the new term will become the new ending point of the sum? $\endgroup$
    – user537153
    Apr 14, 2018 at 1:38
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    $\begingroup$ @JamieCorkhill There is no standard way to do it, because it doesn't matter how you do it. They both mean the same thing. $\endgroup$
    – Ovi
    Apr 14, 2018 at 1:39
  • $\begingroup$ @Alex D I think you misunderstood the question; the OP is not asking why when you add $k^2$, $1^2 + 2^2 + \cdots + (k-1)^2$ becomes $1^2 + 2^2 + \cdots + (k-1)^2 + k^2$. They're asking why it becomes $1^2 + 2^2 + \cdots + k^2$ as opposed to $1^2 + 2^2 + \cdots+ (k-1)^2+k^2$ $\endgroup$
    – Ovi
    Apr 14, 2018 at 1:42
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    $\begingroup$ @JamieCorkhill Only if the new term immediately succeeds the previous ending point, like so: $$k^{2}+\sum^{k-1}_{n=1}n^{2}=\big( 1^{2}+2^{2}+...+(k-1)^2\big)+k^{2}=1^{2}+2^{2}+...+k^{2}=\sum^{k}_{n=1}n^{2}.$$ Here, the square of the (k-1) th term is still there. It is just hiding inside the ellipsis points. If we would have added $(k+1)^{2}$ instead of $k^{2}$, we would have: $$\begin{align}(k+1)^{2}+\sum^{k-1}_{n=1}n^{2}&=\big( 1^{2}+2^{2}+...+(k-1)^2\big)+(k+1)^{2}\\&=1^{2}+2^{2}+...+(k-1)^2+(k+1)^{2}\\&= \left(\sum^{k+1}_{n=1}n^{2}\right)-k^{2}\\&\not=\sum^{k+1}_{n=1}n^{2}\end{align}.$$ $\endgroup$
    – Alex D
    Apr 14, 2018 at 2:37

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