1. A. According to the expression a_n=4*a_n-1, each term after a1 is four times the previous term. The first term is -7 as given, 2nd term should be -7*4=-28, 3rd term is -28*4=-112, ... A is the correct answer.
2. B. The sequence is -13, -8, -3, 2... It's obvious that each term is equal to the previous term plus 5. This is an arithmetic sequence with initial term -13 and common difference 5. We know a1=-13, so a_n=-13+5*(n-1). The answer is B.
3. A. We are given a15=-53, a16=-5. The common difference of the arithmetic sequence is -5-(-53)=48. The formula for a_n term is a1+48*(n-1). We know that a15=-13; plug in n=15, a15=-53=a1+48*(15-1), a1=-725. So a_n=-725+48*(n-1).
4. Diverge. We are given a few terms, 11, 44, 176, 704... Observe that each term is four times the previous one. 11*4=44, 44*4=176, 176*4=704... This is a geometric series with common ratio>1. You can keep multiplying by 4 and the series goes to infinity, so it diverges.
5. D. We have -4, -16, -64, -256... Same as above, each term is four times the previous one. The initial term is a1=-4. The common ratio d=4. So a_n=a1*d^(n-1)=-4*4^(n-1)=-4^n. (D).
6. The answer is A. a2=-2, a5=16. Suppose the common ratio is D. a_n=a1*d^(n-1). a2=a1*d; a5=a1*d^4. Plug in a2 and a5: -2=a1*d, 16=a1*d^4. 16/-2=d^3=-8, d=-2, a1=1. So a_n=1*(-2)^(n-1).
7. B. We are given the sequence 4, -24, 144,... Each term is -6 times the previous one. The first term a0=4, the n^th term a_(n-1) is a1*d^n=4*(-6)^n. To express the sum, we simply have to use the sigma notation and sum 4*(-6)^n from n=0 to infinity. The answer is B.
8. D. We are given -3 + 6 + 15 + 24... 132. Each term is equal to the previous one plus 9. First term a0=-3, n^th term a_n-1 is -3+9*n. The last term is 132. 132 =-3+9n, n=15. So we have to sum -3+9n from n=0 to n=15.
9. B. 343 + 512 + 729 + 1000+... 343=7^3, 512=8^3, 729=9^3, 1000=10^3. This is a sequence of perfect cubes. Therefore, the sum is n^3 from n=7 to infinity. (The initial term is 343=7^3).
10. B. We are given 3, 5, 7, 9, ... 21. The common difference is 2. There are (21-3)/2+1=10 terms. The initial term a1=3, and last term is a10=21. The sum is (a1+a10)*10/2=(3+21)*10/2=120.
11. C. 4/3, 16/3, 64/3, 256/3, 1024/3. Each term is four times the previous one. This is a geometric series with initial term a1=4/3 and common ratio r=4. 1024/3 is the 5th term of the sequence. So sum=a1*(1-r^n)/(1-r)=4/3*(1-4^5)/(1-4)=-4/9*-1023=1364/3.
12. B. 10,12,14,... This is an arithmetic sequence. a1=10, and common difference d=2. There are 20 terms (20 rows). a20=a1+d*(n-1)=10+2*(20-1)=48. So the sum S=(a1+an)*n/2=(10+48)*20/2=580.
13. 10 + 20 + 30 + ... + 10n = 5n(n + 1). When n=1, this expression is true, since 10=5*1*(1+1). Suppose when n=k, this statement is true, then when n=k+1, the left side is 10+...+10n+10(n+1), the right side is 5(n+1)(n+2). The left side adds 10(n+1) compared to the previous one. The right side adds 5(n+1)(n+2)-5n(n+1)=5(n+1)(n+2-n)=10(n+1). So the statement holds true.
14. The height at week 0 is a0=300 (initial height). Common difference is 4.2 (weekly increment). a_n=300+4.2n. At week n, the height of the tree is 300+4.2*n centimeters.