write the finite series -1+2+7+14+23+...+62 in summation notation
term(1) = 1^2 -2 = -1
term(2) = 2^2 - 2 = 2
term(3) = 3^2 - 2 = 7
term(4) = 4^2 - 2 = 14
...
term(n) = n^2 - 2
62 = 8^2 - 2 , so we have 8 terms
sigma (n^2 - 2) for 1 to n
= n(n+1)(2n+1)/6 - 2n
recalling that sum n^2 = n(n+1)(2n+1)/6
checking for sum(5)
= 5(6)(11)/6 - 10 = 55-10 = 45
-1+2+7+14+23 = 45
Sure, here's the summation notation for the given series:
∑ (n² + 1)
Where the sum ranges from n = 1 to n = 8.
Now, don't worry, I'm not clowning around when it comes to math! Just having a bit of fun while we dive into these numbers.
To write the finite series -1 + 2 + 7 + 14 + 23 + ... + 62 in summation notation, we need to find the general formula for the terms of the series.
Observing the pattern, we can see that each term is obtained by adding a specific number to the previous term. The series starts with -1 and increases by 3, then by 5, then by 7, and so on.
Therefore, the general term for this series can be written as:
a_n = -1 + (n-1)(3 + 2(n-2))
where n represents the position of the term.
Now, we can express the finite series using summation notation:
Summation notation: Σ( -1 + (n-1)(3 + 2(n-2)) ), from n = 1 to 11
This denotes the sum of the terms of the series from n = 1 to n = 11.
To express the finite series -1 + 2 + 7 + 14 + 23 + ... + 62 in summation notation, we need to determine the pattern governing the terms.
By observing the series, we can see that each term is obtained by adding an arithmetic progression starting from 3, where the first term is 3, the common difference is 5, and the number of terms is increasing by 1.
So, the general formula for the terms in this series can be represented as:
a_n = 3 + 5(n-1),
where 'n' denotes the position of the term in the series.
Now, we can write the summation notation for this series using the formula:
∑(3 + 5(n-1)), with the lower limit being 1 and the upper limit being 12.
Therefore, the summation of the series can be written as:
∑(3 + 5(n-1)), from n = 1 to n = 12.