dave13 wrote:

Harley1980 wrote:

Bunuel wrote:

For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

\(B_n = B_{n-1} + 5\) if n is odd and greater than 1;

\(Bn = -B_{n-1}\) if n is even;

\(B_1 = 3\)

What is the sum of the first 65 terms in the sequence?

(A) –5

(B) 0

(C) 3

(D) 5

(E) 8

B1 = 3

B2 = -3

B3 = 2

B4 = -2

B5 = 3 and from this moment we see that pattern will be repeat each four numbers

65 / 4 = 4*16 + 1

so B65 will be equal to 3 (16 repetions + 1 more calculation)

Answer is C

how do we get these values ?

B2 = -3

B3 = 2

B4 = -2

hi

pushpitkc can you explain pls

dave13 , you will have to deal with my pinch-hitting for

pushpitkc I can't quite tell what part does not make sense, so I may give you too much info. The way that the instructions are written may be throwing you off.

This problem is like a recipe for a sequence. The "recipe" tells us how to calculate each term.

The "ingredients" are terms with even- and odd-numbered subscripts. Even- and odd-numbered terms have different rules.

We specify each term by using a subscript, \(B_{n}\). The first term is \(B_1\), where \(n=1\). Second term is \(B_2\), where \(n=2\)

As mentioned, if the subscript is an odd number, we follow one rule. If subscript is even, we follow another rule.

Rule (1): IF \(n\) in \(B_{n}\) is ODD and greater than 1, use this rule: \(B_{n}=B_{(n-1)}+5\)

That rule means: to calculate the value of THIS term, use the value of the preceding term and add 5 to it.

\((n-1)\) means "use the term with a subscript that is 1 less than this term's subscript"

If we were calculating \(B_5\), we would plug in the value of \(B_4\).(then add 5) \(B_5\):

\(B_{(n-1)}=B_{(5-1)}=B_4\)

\(B_5\) thus will equal \(B_4+5\)

Rule (2) IF \(n\) in \(B_{n}\) is EVEN, use this rule: \(B_{n} = -(B_{(n-1)})\)

Meaning: take the value of the preceding term (n-1), and negate it.

1) Start calculating sequence terms. We need a pattern. Given:

\(B_1=3\)

Next term, \(B_2\)? Use rule for evens. Negate the preceding term

\(B_2=-(B_{(n-1)})=-(B_{(2-1)})=-(B_1)=-(3)\)That is, \(B_2=-3\)

\(B_3\)? Use the rule for odds

\(B_1=3\)

\(B_2=-(B_{(2-1)})=-(B_1)=-3\)

\(B_3=(B_{(n-1)}+5)=(B_2+5)=(-3+5)=2\)

\(B_4=-(B_3)=-(2)=-2\)

\(B_5=(B_4+5)=(-2+5)=3\)

\(B_6=-(B_5)=-(3)=-3\)

\(B_7=(B_6+5)=(-3+5)=2\)

\(B_8=-(B_7)=-(2)=-2\)

We have a pattern that repeats in cycles of 4 terms: \(3, -3, 2, -2\)

2) Cyclicity

We need the sum of "the first 65 terms." We will use the cycles of four terms to find that sum. Cyclicity in this case is like integer powers and units digits cyclicity.

We need to know where, in the cycle of four terms, \(B_{65}\) will "fall."

To find where \(B_{65}\) falls in the cycle of four terms:

Divide the subscript "65" by the cyclicity of "4" Remainder? Match the remainder to a subscript. The term whose subscript equals the remainder has the same value as \(B_{65}\)

No remainder? Then \(B_{65}\) has the same value as \(B_4\) (the term is a multiple of 4)

\(\frac{65}{4}=16\) + remainder

1Thus \(B_{65}=B_1=3\)

3) SUM??

If we were literally to sum all the terms, the first 8 would be

\(3+(-3)+2+(-2)+(3)+(-3)+2+(-2)=\)No thanks. We do not need to sum all the terms.

Every 4 terms, the sum of terms = 0: \(3+(-3)+2+(-2)=0\)

\(B_{65}\) had a "remainder 1." One term before it, \(B_{64}\) was the last term in a cycle of four numbers

So up to the 64\(^{th}\) term, from \(B_1\) to \(B_{64}\), the sum is 0. (16 cycles of four terms; every 4 terms= 0).

\(B_{65}=3\)

\((0+0+0+0+0+0. . . +3)=3\)

The sum of the first 65 terms \(= 3\)

Hope that helps.

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