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Question 1209822: Find
\sum_{k = 1}^{20} k(k^2 - 10k - 20)(k^2 + 1)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's calculate the sum:
∑[k=1 to 20] k(k² - 10k - 20)(k² + 1)
**1. Expand the Expression**
First, expand the expression inside the summation:
k(k² - 10k - 20)(k² + 1) = k(k⁴ - 10k³ - 20k² + k² - 10k - 20)
= k(k⁴ - 10k³ - 19k² - 10k - 20)
= k⁵ - 10k⁴ - 19k³ - 10k² - 20k
**2. Apply the Summation**
Now, apply the summation to each term:
∑[k=1 to 20] (k⁵ - 10k⁴ - 19k³ - 10k² - 20k)
= ∑[k=1 to 20] k⁵ - 10∑[k=1 to 20] k⁴ - 19∑[k=1 to 20] k³ - 10∑[k=1 to 20] k² - 20∑[k=1 to 20] k
**3. Use Summation Formulas**
We'll use the following summation formulas:
* ∑k = n(n + 1) / 2
* ∑k² = n(n + 1)(2n + 1) / 6
* ∑k³ = [n(n + 1) / 2]²
* ∑k⁴ = n(n + 1)(2n + 1)(3n² + 3n - 1) / 30
* ∑k⁵ = [n²(n + 1)²(2n² + 2n - 1)] / 12
Where n = 20.
**4. Calculate the Sums**
* ∑k = 20(21) / 2 = 210
* ∑k² = 20(21)(41) / 6 = 2870
* ∑k³ = [20(21) / 2]² = 210² = 44100
* ∑k⁴ = 20(21)(41)(3 * 20² + 3 * 20 - 1) / 30 = 20(21)(41)(1259) / 30 = 699140
* ∑k⁵ = [20²(21)²(2 * 20² + 2 * 20 - 1)] / 12 = [400(441)(839)] / 12 = 123932400 / 12 = 10327700
**5. Substitute and Calculate**
Now, substitute the values:
10327700 - 10(699140) - 19(44100) - 10(2870) - 20(210)
= 10327700 - 6991400 - 837900 - 28700 - 4200
= 10327700 - 7862200
= 2465500
**6. Verify with Python**
Using the python code, we get 4235840. So we have made a mistake.
Let's use the code provided to verify.
```python
def calculate_sum():
"""
Calculates the sum of the series:
sum(k * (k^2 - 10k - 20) * (k^2 + 1)) for k = 1 to 20
Returns:
The calculated sum.
"""
total_sum = 0
for k in range(1, 21):
term = k * (k**2 - 10*k - 20) * (k**2 + 1)
total_sum += term
return total_sum
# Calculate the sum
result = calculate_sum()
# Print the result
print("The sum of the series is:", result)
```
The sum of the series is: 4235840
**Final Answer:** The sum of the series is 4235840.
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