What is #sum_(i=1) ^4 (4i-2)#?

1 Answer
Sep 4, 2016

#sum_(i = 1)^4 (4i - 2) = 32#

Explanation:

Step 1: Determine the number of terms

In #sum_k^n#, the number of terms is given by #n - k + 1#.

So, in this expansion, there are #4 - 1 + 1 = 4# terms.

Step 2: Determine the first term, #a# and the common difference, #d#

We do this by inputting the first two values of #i#, #1 and 2#, into the general term, given by #4i - 2#.

#i = 1 -> 4 - 2 = 2#

#i = 2 -> 4(2) - 2 = 6#

The common difference is #4# and the first term is #2#.

We can now apply the formula for sum of an arithmetic series to evaluate the sigma.

#s_n = n/2(2a + (n - 1)d)#

#s_4 = 4/2(2(2) + (4- 1)4)#

#s_4 = 2(4 + 12)#

#s_4 = 2(16)#

#s_4 = 32#

So, #sum_(i = 1)^4 (4i - 2) = 32#.

Hopefully this helps!