1. Write the following in expanded for then find the sum
a) ^7(Sigma sign) k=0 (k=1)(k+2)
b) ^7(Sigma sign) k=4 3k
Note: The sigma sign does not show up when pasted here. The 7 is on top of the sigma sign and the k= 0 or k=2, respectively, lies below the sigma sign.
2. Express the following series using sigma notation:
a) (1 X 0)/2 + (2 X 1)/3 + (3 X 2)/4 + ........ (10 X 9)/11
b) (1 X 2)/2 + (2 X 3)/4 + (3 X 4)/6 + .......... (8 X 9)/16
3. Use the Binomial Theorem to expand the following equation:
a) (2p - 3q)^4
b) (2p + 1)^4
4. Find the fifth term in the expansion of (4x + 2y)^7
5. Find the sum of the following series, or state that the sum does not exist
a) 12 + 8 + 16/3 + .......
b) 1/12 + 1/2 + 3 + .........
6. Find the limit of the following or state that the limit does not exist
a) (2n - 1)/n^3
b) ......as n approches infinity of (2n^2)/(n^2 - 1)
Solution Preview
1. ∑_(k=0)^7▒(k+1)(k+2)
First, since this is summation from k=0 to k=7,
We could transform the expression into the following:
∑_(k=0)^7▒(k+1)(k+2) =(0+1)*(0+2)+(1+1)*(1+2)+(2+1)*(2+2)+(3+1)*(3+2)+(4+1)*(4+2)+(5+1)*(5+2)+(6+1)*(6+2)+(7+1)*(7+2)=1*2+2*3+3*4+4*5+5*6+6*7+7*8+8*9=240
2. ∑_(k=4)^7▒〖3k=3*4+3*5+3*6+3*7=66〗
3. a)(1*0)/2+(2*1)/3+(3*2)/4+⋯+(10*9)/11
First, in the first value, for the two numbers in the numerators, they are 1 and 2 less than denominator respectively.
Therefore, if k starts from 0, the first number a0=(0+1)*0/(0+2) or (k+1)k/(k+2) in general. ...
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