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Question 1118204: 10. Evaluate the following. Show your work using concepts from this unit.
a) 1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203
b) 300 - 299 + 298 - 297 + ... + 100 - 99
Found 4 solutions by math_helper, greenestamps, ikleyn, stanbon:
Answer by math_helper(2461)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
I don't know what concepts are in the unit you are studying; the following is the way I would solve these problems.
a) 1 - 3 + 5 - 7 + ... + 201 - 203
To count the number of terms, we can ignore the signs and look at the sequence as an arithmetic sequence with common difference 2. The number of terms is  .
Then, taking into account the alternating signs, we can group the 102 terms into 51 pairs, in each of which the sum is -2. So the sum of this sequence is 51(-2) = -102.
You could also determine the sum less formally by looking at the pattern of the partial sums:
1-3 = -2
-2+5 = 3
3+-7 = -4
4+9 = 5
5+-11 = -6
...
Whenever the last term added is -n, the sum of the sequence is -(n+1)/2. Since the last term in the sequence is -203, the sum of the sequence is -(203+1)/2 = -102.
b) 300 - 299 + 298 - 297 + ... - 100 + 99
I won't go through the details; the process is exactly the same.
Ignoring signs to count the number of terms, we find there are 202 of them; taking into account the alternating signs, the 202 terms can be seen as 101 pairs, in each of which the sum is 1; then the sum of the sequence is 101*1 = 101.
And again you could get that result by looking at the pattern of the partial sums -- for each two terms in the sequence, the sum of the terms increases by 1, making the sum of the sequence (202/2)*1 = 101*1 = 101.
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
a) 1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203
Group the numbers in pairs, using parentheses:
1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203 =
= (1-3) + (5-7) + (9-11) + . . + (201-203) =
= -2 + (-2) + (-2) + . . . + (-2) = 51 times (-2) = 51*(-2) = -102.
b) 300 - 299 + 298 - 297 + ... + 100 - 99
The same idea works: Group the numbers in pairs, using parentheses
300 - 299 + 298 - 297 + . . . + 100 - 99 =
= (300-299) + (298-297) + . . . + (100-99) =
= 1 + 1 + . . . + 1 = 101 times 1 = 101*1 = 101.
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Actually, this problem is typical one of Math circles for students who are interested in Math.
These students might be of 5-6-7 grades, who formally don't know arithmetic progressions yet, but have a curious and active mind.
It is also good as an entertainment problem for aged people, who just forgot long time ago what an arithmetic progression is
(or even never knew it !), but (again !) have a curious and active mind.
* * * Good training tool for any age's person who does care about healthiness of his (or her) mind. * * *
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! 10. Evaluate the following. Show your work using concepts from this unit.
a) 1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203
= 1+5+9+13...+201 -[3+7+11+15+...+203]
= 1 + 4+1 + 2*4+1 + 3*4+1 +...+50*4+1 - [4-1 + 2*4-1 + 3*4-1+...+51*4-1]
1st Series:: a(1) = 1 ; d = 4 ; n = 51
Sum of 1st Series:: (51/2)(1 + 201) = 51*101 = 5151
2nd Series:: a(1) = 3 ; d = 4 ; n = 51
Sum of 2nd Series:: (51/2)(3+203) = 51*103 = 5253
Ans: 5151-5253 = -2
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Cheers,
Stan H.
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