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What is this equal to?

What is this equal to?-example-1
User Lakremon
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1 Answer

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ANSWER


\sum_(n=1) ^(32) (4n + 1) = 2144

Step-by-step explanation

The given series is


\sum_(n=1) ^(32) (4n + 1)

The first term in this series is


a_1=4(1) + 1 = 5

The last term is


l = 4(32) + 1 = 129

The sum of the first n terms is


S_n= (n)/(2) (a + l)

The sum of the first 32 terms is


S_ {32} = (32)/(2) (5 + 129)


S_ {32} =16 * 134


S_ {32} =2144

Therefore,


\sum_(n=1) ^(32) (4n + 1) = 2144

User Lafleur
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