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Airida [17]
3 years ago
7

Compute 1 + 2 + 3 + 4 + ... +48 +49 + 50.​

Mathematics
2 answers:
Maurinko [17]3 years ago
6 0
Answer: 1275


Explanation:
ad-work [718]3 years ago
4 0

Answer:

1275

Step-by-step explanation:

This is an arithmetic series. The formula for this is Sₙ = (n/2)(a₁ +aₙ)d. a₁ is the first term, so here it is 1, and aₙ is the nth term, or the last term, which is 50 here, but we don't know n.

Now we have to use the equation for an arithmetic sequence to solve for n. A sequence would just be if there was not a last number and it went on forever. that equation looks like aₙ = a₁ + (n - 1)d. Now the only new variable is d, which is the common difference. You can find that by subtracting one term from the term before it, like 2-1 = 1, so d is 1.

We can now solve for n by plugging our numbers into the second equation, so 50 = 1 + (n - 1)1, we can distribute the 1 and to (n-1) and get 50 = 1 + n - 1. Now the ones will cancel and we are left with n = 50

Finally we can plug everything into our original equation and find Sₙ = (50/2)(1+50), which simplifies to Sₙ = 25(51), and Sₙ = 1275.

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