Question

PLS HELP I'M SOO CONFUSED

Consider the arithmetic senes 336+ 332+328+....+4 Numeric Response 1 The number of terms in the series above is (Record your answer in the numerical-response section below) Your answer DCC Numeric Response 2. The sum of the series above is (Record your answer in the numerical-response section below) ​

Answer 1

The first term in the series is 336. The second term is obtained by subtracting 4 from the first term; subtract 4 from that to get the third term; and so on. Then the n-th term in the series is

336 - 4 (n - 1)

or

340 - 4n

The last term in this series is 4, so we solve for n :

340 - 4n = 4

336 = 4n

n = 336/4 = 84

The sum of the series is then

`\displaystyle\sum_(n=1)^(84)(340-4n) = 340\sum_(n=1)^(84)1 - 4 \sum_(n=1)^(84)n`

Recall that

`\displaystyle \sum_(n=1)^(N) 1 = N`

`\displaystyle \sum_(n=1)^N n = \frac{N(N+1)}2`

Then the sum we want is

`\displaystyle\sum_(n=1)^(84)(340-4n) = 340*84 - 4*\frac{84*85}2 = \boxed{14,280}`