Difference between revisions of "PaperMath’s sum"
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==Proof== | ==Proof== | ||
− | We will first prove a easier variant of | + | We will first prove a easier variant of PaperMath’s sum, |
<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> | ||
− | This is the exact same as | + | This is the exact same as |
<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> | ||
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Notice that <math>9(\underbrace {22\dots}_{n})=2(\underbrace {99\dots}_{n})</math> | Notice that <math>9(\underbrace {22\dots}_{n})=2(\underbrace {99\dots}_{n})</math> | ||
− | Substituting this into <math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})</math> yields | + | Substituting this into <math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})</math> yields <math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})</math> |
− | <math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})</math> | ||
Adding <math>1</math> on both sides yields | Adding <math>1</math> on both sides yields | ||
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As you can see, | As you can see, | ||
− | + | ||
<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> | ||
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<math>x^2\sum_{i=0}^{2n} {10^i}=(x \sum_{j=0}^n {(3 \times 10^j)})^2 + x^2\sum_{k=0}^n {(2 \times 10^k)}</math> | <math>x^2\sum_{i=0}^{2n} {10^i}=(x \sum_{j=0}^n {(3 \times 10^j)})^2 + x^2\sum_{k=0}^n {(2 \times 10^k)}</math> | ||
− | Which proves | + | Which proves PaperMath’s sum |
==Problems== | ==Problems== |
Revision as of 15:00, 22 January 2024
Contents
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
Problems
2018 AMC 12A Problem 25
For a positive integer and nonzero digits , , and , let be the -digit integer each of whose digits is equal to ; let be the -digit integer each of whose digits is equal to , and let be the -digit (not -digit) integer each of whose digits is equal to . What is the greatest possible value of for which there are at least two values of such that ?
Notes
AntandMonkeyDoctor‘s sum was discovered by the aops user AntandMonkeyDoctor, as the name implies. It was stolen from AntandMonkeyDoctor.