Question

Use S = n2 to find the sum of 1 + 3 + 5 + . . . + 499

Answer 1

just a sec

Answer 2

a(n)=1+2(n-1)
499=1+2(k-1)
498=2(k-1)
249=k-1
k=248

Answer 3

S = 1 + 3 + 5 + . . . + 499
This is an arithmetic series with first term \(a_1 = 1\), common difference \(d = 2\) and the last term 499. First, find how many terms there are in the series. The nth term of an arithmetic sequence is given by the formula: \(a_n = a_1 + (n-1)d\). Therefore,
\[499 = 1 + (n-1)*2 \\
498 = 2(n-1) \\
249 = n - 1 \\
n = 250
\]The sum of the series is given by the formula:
\(S_n = (a_1 + a_{last})/2 * n = (1 + 499)/2 * 250 = ~?\)