Difference between revisions of "PaperMath’s sum"

Revision as of 21:26, 1 September 2024

PaperMath’s sum

Papermath’s sum states,

$\sum_{i=0}^{2n-1} {(x^2 \times 10^i)}=(\sum_{j=0}^{n-1}{(3x \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2x^2 \times 10^k)}$

Or

$x^2\sum_{i=0}^{2n} {10^i}=(x \sum_{j=0}^n {(3 \times 10^j)})^2 + x^2\sum_{k=0}^n {(2 \times 10^k)}$

For all real values of $x$ , this equation holds true for all nonnegative values of $n$ . When $x=1$ , this reduces to

$\sum_{i=0}^{2n-1} {10^i}=(\sum_{j=0}^{n -1}{(3 \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2 \times 10^k)}$

Proof

First, note that the $x^2$ part is trivial multiplication.

Simple Proof

$(10^{2n}-1)/9 = 9((10^{n}-1)/9)^2 + 2(10^n -1)/9$

Papermath's proof

We will first prove a easier variant of Papermath’s sum,

$\sum_{i=0}^{2n} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}$

This is the exact same as

$\sum_{i=0}^{2n} {10^i}=(\sum_{j=0}^n {(3 \times 10^j)})^2 + \sum_{k=0}^n {(2 \times 10^k)}$

But everything is multiplied by $9$ .

Notice that this is the exact same as saying

$\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})$

Notice that $9(\underbrace {22\dots}_{n})=2(\underbrace {99\dots}_{n})$

Substituting this into $\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})$ yields $\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})$

Adding $1$ on both sides yields

$10^{2n}= (\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})+1$

Notice that $(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})+1=(\underbrace {99\dots}_{n}+1)^2=(10^n)^2=10^{2n}$

As you can see,

$\sum_{i=0}^{2n} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}$

Is true since the RHS and LHS are equal

This equation holds true for any values of $n$ . Since this is true, we can divide by $9$ on both sides to get

$\sum_{i=0}^{2n} {10^i}=(\sum_{j=0}^n {(3 \times 10^j)})^2 + \sum_{k=0}^n {(2 \times 10^k)}$

And then multiply both sides $x^2$ to get

$\sum_{i=0}^{2n} {(x^2 \times 10^i)}=(\sum_{j=0}^n {(3x \times 10^j)})^2 + \sum_{k=0}^n {(2x^2 \times 10^k)}$

Or

$x^2\sum_{i=0}^{2n} {10^i}=(x \sum_{j=0}^n {(3 \times 10^j)})^2 + x^2\sum_{k=0}^n {(2 \times 10^k)}$

Which proves Papermath’s sum

Problems

AMC 12A Problem 25

For a positive integer $n$ and nonzero digits $a$ , $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$ ?

$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

Notes

Papermath’s sum was named by the aops user Papermath. The name is not widely used.

@@ Line 1: / Line 1: @@
 == PaperMath’s sum==
-This is a summation identities for decomposition or reconstruction of summations. Papermath’s sum states,
+ 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-1} {(x^2 \times 10^i)}=(\sum_{j=0}^{n-1}{(3x \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2x^2 \times 10^k)}</math>
 Or
@@ Line 10: / Line 10: @@
 For all real values of <math>x</math>, this equation holds true for all nonnegative values of <math>n</math>. When <math>x=1</math>, this reduces to
-<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-1} {10^i}=(\sum_{j=0}^{n -1}{(3 \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2 \times 10^k)}</math>
 ==Proof==

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