Difference between revisions of "PaperMath’s sum"
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− | == | + | == PaperMath’s sum== |
− | This is a summation identities for decomposition or reconstruction of summations. | + | This is a summation identities for decomposition or reconstruction of summations. Papermath’s sum states, |
<math>\sum_{i=0}^{2n} {(x^2 \times 10^i)}=(\sum_{j=0}^n {(3x \times 10^j)})^2 + \sum_{k=0}^n {(2x^2 \times 10^k)}</math> | <math>\sum_{i=0}^{2n} {(x^2 \times 10^i)}=(\sum_{j=0}^n {(3x \times 10^j)})^2 + \sum_{k=0}^n {(2x^2 \times 10^k)}</math> | ||
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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> | ||
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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== | ||
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==Notes== | ==Notes== | ||
− | + | Papermath’s sum was discovered by the aops user Papermath, as the name implies. | |
==See also== | ==See also== |
Revision as of 19:35, 26 February 2024
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
Problems
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
Papermath’s sum was discovered by the aops user Papermath, as the name implies.