phelps wrote:In the sequence 1, 2, 4, 8, 16, 32, ..., each term after the first is twice the previous term.
What is the sum of the 16th, 17th, and 18th terms in the sequence?
A. 218 ...its 2 exponent 18
B. 3(217)...its 3* (2 exponent 17)
C. 7(216) ...its 7*( 2 exponent 16)
D. 3(216) ...its 3* ( 2exponent 16)
E. 7(215)...its 7* ( 2 exponent 15)
Hi,
since the first term is 1, and since we double to find the next term, we can quickly see that the nth term of the sequence is simply 2^(n-1). In other words:
term1 = 2^0
term2 = 2^1
term3 = 2^2
and so on.
So, the 16th, 17th and 18th terms will be:
2^15, 2^16 and 2^17.
Next, we need to take the sum of these terms. We have a simple rule for multiplying exponents with the same base:
x^a * x^b = x^(a+b)
for dividing exponents with the same base:
x^a/x^b = x^a-b)
and for raising a power to another power:
(x^a)^b = x^(a*b)
but there's no simple rule for adding or subtracting exponents with the same base. The only way we can add or subtract is if both the base and the exponent are equal - and we do so simply by adding or subtracting the coefficients. For example:
3(x^4) + 2(x^4) = 5(x^4).
So, if we want to add our three terms, we need to equalize the exponents. To do so, we need to factor down the bigger exponents to match the smallest one.
Since 2^15 is the smallest, we leave that alone.
2^16 = 2^1 * 2^15 = 2(2^15)
2^17 = 2^2 * 2^15 = 4(2^15)
Now we can sum our terms:
2^15 + 2^16 + 2^17 = 1(2^15) + 2(2^15) + 4(2^15) = 7(2^15)... choose (E).
