Question 3709: I am having a very hard time understanding college algebra. Can anyone help me solve these problems.
The first 3 problems have to do with sequences.
1. 5,3,7,1,9,-1 Find a9 then find sum of 9
2. 12,9,6, What is 25n, and what is sum of 25 (s25)
3. 4,8,16 Find a12 and find sum of 12 (s12)
4. (x+y)^7 Expand
5. (2x+3y)^4 Expand
6. (3x-2y)^5 Expand On this last one he gave us a hint but i do not understand. (x-y)=[(x)+(-y)] Can anyone please help!!!!!!!!!!!!!!!!!!!
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! I only solve the questions about seq.
1. 5,3,7,1,9,-1 Find a9 then find sum of 9
Sol: a1 =5, a2 = a1-2, a3=a2+4,a4=a3-6,a5=a4+8,...
In general, we see that an= a(n-1)+ (-1)^(n-1) 2(n-1) for n >= 2.
And,so a7 = -1 + 2*6 = 11, a8 = 11 - 2*7 = -3.
Hence, a9 = -3 + 2*8 = 13.
s9 = s1+s2+...+a9 = 5+3+7+1+9-1+11-3+13 = 25 + 24 - 4 = 45.
In general , sn = a1* n = 5n if n is odd.
and s2n = 10n - 2*n for n >= 1.
2. 12,9,6, What is 25n, and what is sum of 25 (s25)
Sol: This is an arithmetic seq with common difference -3.
an= an-1 -3, so an = a1 + (n-1)(-3) = 15 -3n.
So, a25 = 15 -3*25 = -60.
Also sn = n(a1+an)/2 = n(27 - 3n)/2 = 27n/2 -3n^2/2.
So, s25 = (27*25 - 3*25^2)/2 = -600
3. 4,8,16 Find a12 and find sum of 12 (s12)
Sol: This is a geometric seq with common ratio 2.
an= 2an, so an = a1*2^(n-1) = 4* 2^(n-1) for n>=1.
And, sn = a1(1-r^n)/(1-r) = 4(2^n-1) .
We get a12 = 4*2^11 = 8*1024 = 8192.
and s12 = 4(2^12 -1) = 16380.
The rest concerning binomial formula , should be staightforward.
Kenny
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