If (2n + 1) + (2n + 3) + (2n + 5) + ... + (2n + 47) = 5280, then find 1 + 2 + 3 + ... + n.
(2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47) = 5280
There are 24 terms in the list: (2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47)
So, the left-hand side can be rewritten as (2n)(24) + (1 + 3 + 5 + ... + 47) = 5280
We can use the formula: Sum = n(t1 + tn) / 2 to find the sum of 1 + 3 + ... + 47:
Sum = 24(1 + 47)/2 = 24(48)/2 = 576
Equation: (2n)(24) + (576) = 5280
48n + 576 = 5280
48n = 4704
n = 98
Now, to find the sum: 1 + 2 + 3 + ... + 98 using the formula: Sum = n(t1 + tn) / 2
Sum = 98(1 + 98)/2 = 4851.
(2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47) = 5280
There are 24 terms in the list: (2n + 1) + (2n + 3) + (2n + 5) + ... (2n + 47)
So, the left-hand side can be rewritten as (2n)(24) + (1 + 3 + 5 + ... + 47) = 5280
We can use the formula: Sum = n(t1 + tn) / 2 to find the sum of 1 + 3 + ... + 47:
Sum = 24(1 + 47)/2 = 24(48)/2 = 576
Equation: (2n)(24) + (576) = 5280
48n + 576 = 5280
48n = 4704
n = 98
Now, to find the sum: 1 + 2 + 3 + ... + 98 using the formula: Sum = n(t1 + tn) / 2
Sum = 98(1 + 98)/2 = 4851.