Difference between revisions of "PaperMath’s sum"
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<math>\sum_{i=0}^{2n} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}</math> | <math>\sum_{i=0}^{2n} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}</math> | ||
− | + | Is true since the RHS and LHS are equal | |
+ | |||
+ | This equation holds true for any values of <math>n</math>. Since this is true, we can divide by <math>9</math> on both sides to get | ||
<math>\sum_{i=0}^{2n} {10^i}=(\sum_{j=0}^n {(3 \times 10^j)})^2 + \sum_{k=0}^n {(2 \times 10^k)}</math> | <math>\sum_{i=0}^{2n} {10^i}=(\sum_{j=0}^n {(3 \times 10^j)})^2 + \sum_{k=0}^n {(2 \times 10^k)}</math> |
Revision as of 15:33, 8 October 2023
Contents
[hide]PaperMath’s sum
This is a summation identities for decomposition or reconstruction of summations. PaperMath’s sum states,
Or
For all real values of , this equation holds true for all nonnegative values of . When , this reduces to
Proof
We will first prove a easier variant of PaperMath’s sum,
This is the exact same as
But everything is multiplied by .
Notice that this is the exact same as saying
Notice that
Substituting this into yields
Adding on both sides yields
Notice that
As you can see,
Is true since the RHS and LHS are equal
This equation holds true for any values of . Since this is true, we can divide by on both sides to get
And then multiply both sides to get
Or
Which proves PaperMath’s sum
Notes
PaperMath’s sum was discovered by the aops user PaperMath, as the name implies.