SOLUTION: Direction: use one of the formulas for special series to find the sum of the series. 42 ∑ 1 i=1 1. 42 is on top of the symbol, i=1 is on bottom and 1 on right side.

Algebra ->  Sequences-and-series -> SOLUTION: Direction: use one of the formulas for special series to find the sum of the series. 42 ∑ 1 i=1 1. 42 is on top of the symbol, i=1 is on bottom and 1 on right side.       Log On


   

Question 550929: Direction: use one of the formulas for special series to find the sum of the series.
42
∑ 1
i=1
1. 42 is on top of the symbol, i=1 is on bottom and 1 on right side.
2. 5 on top, n on right, n=1 on bottom. the answer is 15 but how?
3. 18 on top, i=1 on bottom, i on right.
4. 20 on top, k on right, k=1 on bottom
5. 6 on top, n^2 on right, n=1 on bottom.
6. 10 on top, i^2 on right, i=1 on bottom
7. 12 on top, i^2 on right, i=1 on bottom
8. 35 on top, k^2 on right, k=1 on bottom
again all these problems are shown in this symbol: ∑
thanks for ur help and happy new year!
please show ur work with each of the problems so i could understand it. thank you very much.
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do 4 of them.  You can do the other 4.  They are similar to these.

1. 42 is on top of the symbol, i=1 is on bottom and 1 on right side. 
sum%281%2Ci=1%2C42%29 = 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = 42

2. 5 on top, n on right, n=1 on bottom. the answer is 15 but how?
sum%28n%2C1=1%2C5%29 = 1+2+3+4+5 = 15 We didn't need a formula but we can use it:
              n%28n%2B1%29%2F2 = 5%285%2B1%29%2F2 = %28%285%29%286%29%29%2F2 = 15

4. 20 on top, k on right, k=1 on bottom
sum%28k%2Ck=1%2C20%29 = 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20

We use the formula  n%28n%2B1%29%2F2 = 20%2820%2B1%29%2F2 = %28%2820%29%2821%29%29%2F2 = 210

8. 35 on top, k^2 on right, k=1 on bottom

sum%28k%5E2%2Ci=1%2C35%29 = 1²+2²+3²+4²+5²+6²+7²+8²+9²+10²+11²+12²+13²+14²+15²+16²+17²+18²+19²+20²+21²+22²+23²+24²+25²+26²+27²+28²+29²+30²+31²+32²+33²+34²+35²

We use the formula n%28n%2B1%29%282n%2B1%29%2F6 = 35%2835%2B1%29%282%2A35%2B1%29%2F6 = %28%2835%29%2836%29%2870%2B1%29%29%2F6 = %28%2835%29%2836%29%2871%29%29%2F6 = 14910.

Edwin