# What is #sum_(k=1) ^n# #(3k+1)#?

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The sum ( \sum_{k=1}^{n} (3k + 1) ) can be calculated by applying the formula for the sum of an arithmetic series. In this case, the formula is ( S = \frac{n}{2}(a_1 + a_n) ), where ( S ) is the sum of the series, ( n ) is the number of terms, ( a_1 ) is the first term, and ( a_n ) is the last term.

For the given series ( 3k + 1 ), the first term ( a_1 ) is obtained by substituting ( k = 1 ), yielding ( a_1 = 3(1) + 1 = 4 ). The last term ( a_n ) is obtained by substituting ( k = n ), yielding ( a_n = 3n + 1 ).

Substituting these values into the formula, we get: [ S = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(4 + 3n + 1) ] [ S = \frac{n}{2}(3n + 5) ]

Therefore, the sum ( \sum_{k=1}^{n} (3k + 1) ) is ( \frac{n}{2}(3n + 5) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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