The arithmetic series 1+5+9+. . . . . +41+45 can be expressed in summation notation as [tex]\sum\limits^{12}_{n=1} {(4n-3)}[/tex]
The given arithmetic series:
1+5+9+. . . . . +41+45
To write the series in summation notation:
Find the common difference (d) of the series.
d = 5 - 1 = 4
Recall the formula for the nth term in an arithmetic sequence:
a(n) = a(1) + (n-1) . d
Substitute a(1) = 1 and d = 4:
a(n) = 1 + (n-1) . 4
a(n) = 4n - 3
Find n that leads to a(n) = 45
a(n) = 4n - 3
45 = 4n - 3
4n = 48
n = 12
Thus, the lower limit of the summation is n =1, and the upper limit is n = 12
Now write in summation notation as:
[tex]\sum\limits^{12}_{n=1} {a(n)}= \sum\limits^{12}_{n=1} {(4n-3)}[/tex]
This summation notation can be further simplified using the properties of summation operator:
[tex]\sum\limits^{12}_{n=1} {(4n-3)}=\sum\limits^{12}_{n=1} {4n}-\sum\limits^{12}_{n=1} {3}[/tex]
[tex]=\sum\limits^{12}_{n=1} {4n}-12\times3[/tex]
[tex]=\sum\limits^{12}_{n=1} {4n}-36[/tex]
Learn more about summation notation here:
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