Math  /  Algebra

QuestionFind the sum. n=372n\quad \sum_{n=3}^{7} 2^{n}

Studdy Solution

STEP 1

What is this asking? We need to find the sum of the powers of 2, starting from 232^3 and going all the way up to 272^7. Watch out! Don't start at 212^1 or 222^2!
We're starting at 232^3.
Also, don't forget to *add* the powers of 2, not multiply them.

STEP 2

1. Expand the Sum
2. Calculate the Powers
3. Add the Powers

STEP 3

Let's **expand** this fancy summation notation into something a little more down-to-earth.
The sigma, \sum, just means "sum up a bunch of stuff." The n=3n=3 at the bottom tells us our **starting value** for nn is **3**, and the 77 at the top tells us where to **stop**!

STEP 4

So, n=372n\sum_{n=3}^{7} 2^{n} means we're summing up 2n2^n for each value of nn from 3 to 7.
Let's write it out: 23+24+25+26+272^3 + 2^4 + 2^5 + 2^6 + 2^7.
Much better!

STEP 5

Now, let's calculate each of these **powers of 2**: 23=222=82^3 = 2 \cdot 2 \cdot 2 = 8 24=2222=162^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 1625=22222=322^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 3226=222222=642^6 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 6427=2222222=1282^7 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 128

STEP 6

Almost there!
Now we just need to **add up** all the values we just calculated: 8+16+32+64+1288 + 16 + 32 + 64 + 128

STEP 7

Let's do this carefully: 8+16=248 + 16 = 24 24+32=5624 + 32 = 5656+64=12056 + 64 = 120120+128=248120 + 128 = 248

STEP 8

So, the **sum** n=372n\sum_{n=3}^{7} 2^{n} is equal to **248**!

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