Fall 1999, test #1 solutions
Solution: 3, -6, 12, -24, 48, -96
Solution: 1, 5/ 3, 7/ 3, 3, 11/ 3, 13/ 3
Solution: 2, 3, 5, 9, 17, 33
Solution: a_{n} = -3 + 2(n-1) or a_{n}
= 2n-5
Solution: a_{n} = 5·2n-1
4 | 8 | 12 | . . . | 192 | 196 | 200 | |
200 | 196 | 192 | . . . | 12 | 8 | 4 | |
Sum | 204 | 204 | 204 | . . . | 204 | 204 | 204 |
There are 50 terms, so the sum is (50·204)/ 2 = 5100.
Method 2: If we use the formula s_{n} = (a_{1}
+ a_{n})/ 2, the calculation is almost the same.
Alternatively, we have s_{n} = 4(1 + 2 + ... + 50) = 4·50·51/
2 = 5100.
1 | 5 | 9 | . . . | 93 | 97 | 101 | |
101 | 97 | 93 | . . . | 9 | 5 | 1 | |
Sum | 102 | 102 | 102 | . . . | 102 | 102 | 102 |
There are 26 terms, so the sum is (26·102)/ 2 = 1326.
Method 2: If we use the formula s_{n} = (a_{1}
+ a_{n})/ 2, the calculation is almost the same.
Alternatively, we have s_{n} = 1 + (1 + 4·1) + (1 +
4·2) + ... + (1 + 25·4).
Thus, s_{n} = 1·26 + 4(1 + 2 + ... + 25) = 26 + 4·25·26/
2 = 1326.
Solution: s = 9·(1 - (2/ 3)^{6})/(1
- (2/ 3)) = 9·(1 - 64/
729)/(1 - (2/ 3)) = 665/ 27 =
24.63
Solution: s = 9·(1 - (2/ 3)^{n})/(1
- (2/ 3)) = 27·(1 - (2/
3)^{n})
Solution: As n tends to infinity, (2/ 3)^{n} tends to 0. So, the value of the sum of the first nterms tends to 27. So, the infinite sum is 27.
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