2025: Geometry

January 23, 2025Chinoiseries2014 IMO, Math Games Leave a comment

Note: Tough if no Geometry knowledge : the R, r are on the same connected line. 2 adjacent circles tangent at 1 point, the tangent at that point of big circle and small circle is perpendicular to both radii R, r

AMS 13/1/2025 Algebra

January 13, 2025Chinoiseries2014 Math Games Leave a comment

Solution :

AMS 12/1/2025: Algebra

January 12, 2025Chinoiseries2014 AI, Education Leave a comment

Solution by DeepSeek:

Perplexity

Thank you for your patience. Let’s carefully evaluate the series

$S = \sum_{x=0}^{\infty} \frac{2x + 1}{2^{x-1}}$

Step 1: Rewrite the Series

We can rewrite the series as follows:

$\displaystyle S = \sum_{x=0}^{\infty} \frac{2x + 1}{2^{x-1}} = \sum_{x=0}^{\infty} (2x + 1) \cdot 2^{-(x-1)}$

Thus, we have:

$\displaystyle S = 2 \sum_{x=0}^{\infty} (2x + 1) \cdot 2^{-x}$

Step 2: Split the Series

Now, we can split the sum into two separate sums:

$\displaystyle S = 2 \left( \sum_{x=0}^{\infty} 2x \cdot 2^{-x} + \sum_{x=0}^{\infty} 2^{-x} \right)$

Step 3: Evaluate Each Sum

Sum of $2^{-x}$

The second sum is a geometric series:

$\displaystyle \sum_{x=0}^{\infty} 2^{-x} = \frac{1}{1 - \frac{1}{2}} = 2$

Sum of $2x \cdot 2^{-x}$

For the first sum, we use the formula for the sum of $x r^x$

$\displaystyle \sum_{x=0}^{\infty} x r^x = \frac{r}{(1 - r)^2}$

Setting $r = \frac{1}{2}$ :

$\displaystyle \sum_{x=0}^{\infty} x \left(\frac{1}{2}\right)^x = \frac{\frac{1}{2}}{\left(1 - \frac{1}{2}\right)^2} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2} = 2$

Thus,

$\displaystyle \sum_{x=0}^{\infty} 2x \cdot 2^{-x} = 2*(2) = 4$

Step 4: Combine Results

Now we can combine both results into our expression for $S$ :

$S = 2 \left(4 + 2\right) = 2 \times 6 = 12$

Final Result

Thus, the value of the series is

$\boxed {S = 12}$

Cici

We want to find the sum of the series \sum_{x=0}^{\infty} \frac{2x + 1}{2^{x-1}}.

Step 1: Rewrite the series

Let’s rewrite the series to make it easier to work with. We can split the fraction:

\sum_{x=0}^{\infty} \frac{2x + 1}{2^{x-1}} = \sum_{x=0}^{\infty} \left( \frac{2x}{2^{x-1}} + \frac{1}{2^{x-1}} \right) = \sum_{x=0}^{\infty} \left( \frac{4x}{2^x} + \frac{2}{2^x} \right) = \sum_{x=0}^{\infty} \frac{4x + 2}{2^x}

Step 2: Separate the series

We can separate the series into two simpler series:

\sum_{x=0}^{\infty} \frac{4x + 2}{2^x} = 4\sum_{x=0}^{\infty} \frac{x}{2^x} + 2\sum_{x=0}^{\infty} \frac{1}{2^x}

Step 3: Evaluate the geometric series

The second series is a geometric series with the first term a = 1 and common ratio r = \frac{1}{2}. Since |r| < 1, the series converges, and its sum is:

2\sum_{x=0}^{\infty} \left(\frac{1}{2}\right)^x = 2 \cdot \frac{1}{1 – \frac{1}{2}} = 2 \cdot 2 = 4

Step 4: Evaluate the remaining series

Let S = \sum_{x=0}^{\infty} x r^x, where r = \frac{1}{2}. We know that \sum_{x=0}^{\infty} r^x = \frac{1}{1-r} for |r| < 1. Differentiating both sides with respect to r, we get:

\sum_{x=1}^{\infty} x r^{x-1} = \frac{1}{(1-r)^2}

Multiplying by r:

\sum_{x=1}^{\infty} x r^x = \frac{r}{(1-r)^2}

Since our series starts at x=0, and the x=0 term is 0, we have:

\sum_{x=0}^{\infty} x r^x = \frac{r}{(1-r)^2}

Plugging in r = \frac{1}{2}:

\sum_{x=0}^{\infty} x\left(\frac{1}{2}\right)^x = \frac{\frac{1}{2}}{(1 – \frac{1}{2})^2} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2

Therefore, 4\sum_{x=0}^{\infty} \frac{x}{2^x} = 4 \cdot 2 = 8

Step 5: Combine the results

Adding the sums of the two series:

4\sum_{x=0}^{\infty} \frac{x}{2^x} + 2\sum_{x=0}^{\infty} \frac{1}{2^x} = 8 + 4 = 12

Answer

Therefore, the sum of the series \sum_{x=0}^{\infty} \frac{2x + 1}{2^{x-1}} is 12.

Deepseek vs Gemini vs Others

January 8, 2025Chinoiseries2014 Math Games Leave a comment

DeepSeek is smarter than other AI in solving the following AMS problem:

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A post shared by 吴老师 (@tomcircle88)

Gemini:

Gemini AI (Google):

Finding x

Understanding the Problem
We are given:

y: An integer
x: Defined as the sum of powers of y from 0 to 7:
x = 1 + y + y² + y³ + y⁴ + y⁵ + y⁶ + y⁷
We need to find the value of x, given that x is a power of a prime number.

Simplifying the Expression
The expression for x is a geometric series. We can simplify it using the formula for the sum of a geometric series:

Sum of a geometric series:
S = a(1 – r^n) / (1 – r)
where:
S is the sum of the series
a is the first term
r is the common ratio
n is the number of terms
In our case:
a = 1
r = y
n = 8
Therefore:
x = 1(1 – y⁸) / (1 – y)
x = (1 – y⁸) / (1 – y)

Finding x when x is a Power of a Prime
To find x when it’s a power of a prime, we can:

Iterate through possible values of y: Start with y = 0 and incrementally increase y.
Calculate x: For each y, calculate the value of x using the simplified formula.
Check if x is a power of a prime:
Determine if x has only one prime factor.
If so, check if x can be expressed as that prime raised to an integer power.
Example (using Python)
import math

def is_power_of_prime(n):
“””
Checks if a number is a power of a prime.

Args:
n: The number to check.

Returns:
True if n is a power of a prime, False otherwise.
“””
if n <= 1:
return False

for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
p = i
break
else:
return n == 1 # 1 is not considered a power of a prime

while n % p == 0:
n //= p

return n == 1

def find_x(max_y):
“””
Finds the value of x for integer values of y up to max_y,
where x = 1 + y + y^2 + … + y^7 and x is a power of a prime.

Args:
max_y: The maximum value of y to consider.

Returns:
A list of values of x that satisfy the conditions.
“””
results = []
for y in range(max_y + 1):
x = (1 – y**8) // (1 – y) if y != 1 else 8
if is_power_of_prime(x):
results.append(x)
return results

Find x for y values up to 10

max_y = 10
x_values = find_x(max_y)

if x_values:
print(f”Values of x for y up to {max_y}: {x_values}”)
else:
print(f”No values of x found for y up to {max_y}.”)

Output:
Values of x for y up to 10: [8]

In this example, the code iterates through y values up to 10 and finds that the only value of x that satisfies the conditions is x = 8 (when y = 1).
Note:

This approach might not be the most efficient for very large ranges of y.
You can further optimize the code by incorporating mathematical properties of prime numbers and powers.
*

Other AI: Copilot, Perplexity, Cici

As a human by “look-see”, we are smarter than AI Gemini by guessing
y=1 to get x = 8 = 2³.

Gemini uses Python to solve – 杀鸡用牛刀！
There other 3 AI tools:
Ceci,
Perplexity，
Copilot
all misled by y≠1
BCOS they use Geometric Series Sum :
${x = \frac {y^{8}-1}{y-1}}$
where y cannot be = 1,
but
x= 1+y+y²+…y⁷
valid for any y.