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# Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the

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Joined: 02 Sep 2009
Posts: 49206
Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the  [#permalink]

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11 Jun 2015, 04:33
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Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the value of K?

(A) 67
(B) 50
(C) 45
(D) 41
(E) 36

Kudos for a correct solution.

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Re: Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the  [#permalink]

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11 Jun 2015, 04:51
2
I used brute force for this question:
S6: 36+30+94=160
S5: 25+25+94=144
S4: 16+20+94=130
S3: 9+15+94=118
S2: 4+10+94=108
S1: 1+5+94=101

S6-S5+S4-S3+S2-S1= 16+12+7=36
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Re: Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the  [#permalink]

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11 Jun 2015, 05:19
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Bunuel wrote:
Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the value of K?

(A) 67
(B) 50
(C) 45
(D) 41
(E) 36

Kudos for a correct solution.

two ways , i would solve it..
1) Sn = n^2 + 5n + 94
K=S6 – S5 + S4 – S3 + S2 – S1
=(6^2-5^2+4^2-3^2+2^2-1^2)+5(6-5+4-3+2-1)+0
=(36-25+16-9+4-1)+5(3)+0=36
2) looking at the choices avail..
we know there are three even terms in +ive and three odd terms in -ive...
so 94 gets cancelled out..
if you add square of three even numbers and subtract square of odd no, it will always be odd..
so n^2 will be odd and 5*n will also be odd, so answer should be even....only B and E are left..

Also if we just concentrate on last digits of n^2 and n.. we can easily see it has to be 6.. only 36 is left
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the  [#permalink]

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15 Jun 2015, 01:51
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Bunuel wrote:
Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the value of K?

(A) 67
(B) 50
(C) 45
(D) 41
(E) 36

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

The expression for K becomes more manageable if we insert parentheses:
K = (S6 – S5) + (S4 – S3 )+ (S2 – S1)

K involves two expressions of the form $$(S_{n + 1} - S_n)$$, so a general rule for $$(S_{n + 1} - S_n )$$ might be helpful. We know that $$S_n = n^2 + 5n + 94$$, so

$$S_{n + 1} = (n + 1)^2 + 5(n + 1) + 94$$
$$S_{n + 1} = (n^2 + 2n + 1) + (5n + 5) + 94$$
$$S_{n + 1} = n^2 + 7n + 100$$

In general, $$(S_{n + 1} - S_n) = (n^2 + 7n + 100) - (n^2 + 5n + 94) = 2n + 6$$.

Applying this rule to the grouped terms in K:
K = (S6 – S5) + (S4 – S3) + (S2 – S1)
= (2 × 5 + 6) + (2 × 3 + 6) + (2 × 1 + 6)
= 16 + 12 + 8
= 36

Alternatively, we could just plug into the Sn term definition, and identify common terms as we work:
K = S6 – S5 + S4 – S3 + S2 – S1= (6^2 + 5(6) + 94) –(5^2 + 5(5) + 94) + (4^2 + 5(4) + 94) – (3^2 + 5(3) + 94) + (2^2 + 5(2) + 94) – (1^2 +5(1) + 94)= (6^2 – 5^2 + 4^2 – 3^2 + 2^2 – 1^2) + 5(6 – 5 + 4 – 3 + 2 – 1)= (36 – 25 + 16 – 9 + 4 – 1) + 5(3)= (21) + (15)= 36

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Re: Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the  [#permalink]

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12 Jul 2015, 04:06
1
Bunuel wrote:
Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the value of K?

(A) 67
(B) 50
(C) 45
(D) 41
(E) 36

Kudos for a correct solution.

Can I just ignore the "+94" in the first equation because we always have a subtraction between Sn - S(n-1)? Like this:

S6-S5 = 6^2 + 5*6 - 5^2 +5*5 =16
S4-S3 = 4^2 + 5*4 - 3^2 + 5*3 = 12
S2-S1 = 2^2 + 2*5 - 1 + 5 = 8
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Re: Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the  [#permalink]

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12 Jul 2015, 04:27
1
reto wrote:
Bunuel wrote:
Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the value of K?

(A) 67
(B) 50
(C) 45
(D) 41
(E) 36

Kudos for a correct solution.

Can I just ignore the "+94" in the first equation because we always have a subtraction between Sn - S(n-1)? Like this:

S6-S5 = 6^2 + 5*6 - 5^2 +5*5 =16
S4-S3 = 4^2 + 5*4 - 3^2 + 5*3 = 12
S2-S1 = 2^2 + 2*5 - 1 + 5 = 8

Yes, you can ignore 94 but only because the number of negative terms = number of positive terms and the coefficients of positive terms (all are 1) = coefficients of negative terms (all are 1).

Had it been : s5-s4+s3-s2+s1 , you could not have 'ignored' 94 as the number of terms for positive vs negative are not equal. So 94 would not have been cancelled.

Also, if it had been: s4- 3s3 + 2s2-s1 , even in this case, you can not ignore 94 (you might still be able to cancel all 94s but you should follow a systematic approach for questions that will provide highest ROI with least scope of error.
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Re: Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the  [#permalink]

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19 Feb 2017, 20:55
Bunuel- how were we supposed to know that s6 meant plug in 6 for n and s5 meant plug in 5 for n? I have not seen this is any of my prep materials. Maybe I am dumb- but I was not able to draw that inference from the question.
Re: Sn = n^2 + 5n + 94 and K = S6 – S5 + S4 – S3 + S2 – S1. What is the &nbs [#permalink] 19 Feb 2017, 20:55
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