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zhenek [66]
4 years ago
9

Express the following sums using sigma notation.a. 1 plus 3 plus 5 plus 7 plus ... plus 99b. 8 plus 11 plus 14 plus ... plus 32c

. 6 plus 10 plus 14 plus ... plus 54d. StartFraction 1 Over 1 times 4 EndFraction plus StartFraction 1 Over 2 times 5 EndFraction plus StartFraction 1 Over 3 times 6 EndFraction plus ... plus StartFraction 1 Over 49 times 52 EndFraction
Mathematics
2 answers:
Annette [7]4 years ago
6 0

Answer:

e following sums using sigma notation.a. 1 plus 3 plus 5 plus 7 plus ... plus 99b. 8 plus 11 plus 14 plus ... plus 32c. 6 plus 10 plus 14 plus ... plus 54d. StartFraction 1 Over 1 times 4 EndFraction plus StartFraction 1 Over 2 times 5 EndFraction plus StartFraction 1 Over 3 times 6 EndFraction plus ... plus StartFraction 1 Over 49 times 52 EndFraction add them

Step-by-step explanation:

jolli1 [7]4 years ago
4 0

Answer:

a. 1+3+5+7+...+99 = Σ \left \{ {{\infty} \atop {n=1}} \right.2n+1,  n∈Z

b. 8+11+14+...+32 = Σ \left \{ {{\infty} \atop {n=1}} \right. 3n+5, n∈Z

c. 6+10+14+...+54 = Σ \left \{ {{\infty} \atop {n=1}} \right. 4n+2, n∈Z

d. \frac{1}{(1)*(4)}+\frac{1}{(2)*(5)}+\frac{1}{(3)*(6)}+...+\frac{1}{(49)*(52)} = Σ \left \{ {{\infty} \atop {n=1}} \right. \frac{1}{(n)(n+3)}, n∈Z

Step-by-step explanation:

So basically you only need to find a possible sequence which represent those list of numbers. Once you have done that you only need to express it as a sum.

You can find more information in the next link:

https://www.mathsisfun.com/algebra/sequences-series.html

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