summation question simple

Question

when you are checking to see if a sum of say $k^2$ from $k=1$ to to $k=n$ is equal to a sum of $(k+1)^2$ from $k=0$ to $n−1$ can someone explain what is going on here. THanks

(looking for a fairly simple way to work the problem without writing out the sums which may help me understand what is going on, )

score 1 · Answer 1 · answered Oct 27 '18 at 19:58

1

Think about what summation notation means. The sum $$\sum_{k=1}^n k^2$$ is evaluated by adding a term for each $k = 1, \ldots, n$, and the term for the given value of $k$ is $k^2$. So, the sum becomes $$\sum_{k=1}^n k^2 = 1^2 + 2^2 + 3^2 + \ldots + n^2.$$ Similarly, $$\sum_{k=0}^{n-1} (k + 1)^2 = (0 + 1)^2 + (1 + 1)^2 + \ldots + (n - 1 + 1)^2 = \sum_{k=1}^n k^2.$$

answered Oct 27 '18 at 19:58

Theo Bendit

50,900

Thanks, how would you work it out without writing the sums out? – Carlos Bacca Oct 27 '18 at 20:01
@JamesAnthony As hamam_Abdullah showed, you can look at it in terms of a variable substitution. But, I think writing out the terms is usually the best way to convince yourself that the sums are equal. – Theo Bendit Oct 28 '18 at 02:37

score 1 · Accepted Answer · answered Oct 27 '18 at 20:02

1

$$S=\sum_{k=1}^nf(k)=\sum_{k=1}^nf(k-i+i)$$

put $k-i=K$ then

$$S=\sum_{K=1-i}^{n-i}f(K+i)$$

you can take $i=1$.

answered Oct 27 '18 at 20:02

hamam_Abdallah

62,951

probably great but doesn't really help me understand the problem – Carlos Bacca Oct 27 '18 at 20:21
@JamesAnthony It is a kind of shift of the index. – hamam_Abdallah Oct 27 '18 at 20:24

summation question simple

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