Step-by-step explanation:
[tex]1.\sum_{i=1}^{5}3i[/tex]
The simplest method is "brute force". Calculate each term and add them up.
∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5)
∑ = 3 + 6 + 9 + 12 + 15
∑ = 45
[tex]2.\sum_{k=1}^{4}(2k)^{2}[/tex]
∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²
∑ = 4 + 16 + 36 + 64
∑ = 120
[tex]3.\sum_{k=3}^{6}(2k-10)[/tex]
∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)
∑ = -4 + -2 + 0 + 2
∑ = -4
4. 1 + 1/4 + 1/16 + 1/64 + 1/256
This is a geometric sequence where the first term is 1 and the common ratio is 1/4. The nth term is:
a = 1 (1/4)ⁿ⁻¹
So the series is:
[tex]\sum_{j=1}^{7}(\frac{1}{4})^{j-1}[/tex]
5. -5 + -1 + 3 + 7 + 11
This is an arithmetic sequence where the first term is -5 and the common difference is 4. The nth term is:
a = -5 + 4(n−1)
a = -5 + 4n − 4
a = 4n − 9
So the series is:
[tex]\sum_{j=1}^{5}(4j-9)[/tex]
The sigma notation allows writing an expression in a compact form using the upper and lower boundaries of the variable, then the actual expression. The solutions to the problems are as follows.
1.)
[tex]\sum_{i= 1}^{5} 3i [/tex] ; values of i ranges from 1 - 5 ;
[tex]\sum_{i= 1}^{5} 3i = [3(1) + 3(2) + 3(3) + 3(4) + 3(5)] = 45 [/tex]
2.)
[tex]\sum_{k= 1}^{4} (2k)^{2} [/tex] ; values of k ranges from 1 - 5 ;
[tex]\sum_{k= 1}^{4} (2k)^{2} [/tex]
= (2×1)² + (2×2)² + (2×3)² + (2×4)²
= 4 + 16 + 36 + 64
= 120
3.)
[tex]\sum_{k= 3}^{6} (2k - 10) [/tex] ; values of k ranges from 3 - 6;
[tex]\sum_{k= 3}^{6} (2k - 10) [/tex]
= (2(3) - 10) + (2(4) - 10) + (2(5) - 10) + (2(6) - 10)
= - 4 - 2 + 0 + 2
= - 4
4.)
The common ratio, r = r2 ÷ r1 = r3 ÷ r2 = r4 ÷ r3 =...
r = 1/4 ÷ 1/1 = 1/4
The nth term of a geometric sequence :
Hence, the series in sigma notation can be expressed thus :
5.)
This is an arithmetic series :
d = -1 - (-5) = -1 + 5 = 4
nth term = a + (n-1)d ; n = number of terms ; a = first term
General expression :
-5 + (n - 1)4
-5 + 4n - 4
4n - 9
[tex]\sum_{j= 1}^{5} (4j - 9) [/tex]
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