Adding $k^2$ to $1^2 + 2^2 + \cdots + (k - 1)^2$.

Question

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.

Answer 1

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.

Answer 2

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:

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

Answer 3

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.