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how many terms of the arithmetic sequence {1,3,5,7,....} will gave a sum of 961?

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Mlcparra16id
When we add odd numbers:
1 = 1 
1 + 3 = 4
1 + 3 + 5 = 9
...
Notice that the sum of n terms is n²

* This is because an odd number is expressed as  2n-1, we would need to get the sum of all 2n-1 substituting the value from 1 to n. (This is summation)
So we would have 2(1+2+3+...+n) - n = 2[n(n+1)/2] - n = n(n+1) - n = n(n+1-1) = n²

So:
Sum of n terms = 961
Sum of n terms = 31² = n²
n = 31

There are 31 terms in the arithmetic sequence
Shinalcantaraid
[tex]Using~the~formula~for~the~sum~of~arithmetic~sequence~you'll~have: \\ S_n=[2A_1 + (n-1)d] \frac{n}{2} \\ Given: \\ S_n = 961 \\ A_1=1 \\ d =2 \\ Substitute~the~given~to~the~formula. \\ 961 = [2(1)+(n-1)(2)] \frac{n}{2} \\ 961 = [2 + 2n - 2 ] \frac{n}{2} \\ 961 = (2n)( \frac{n}{2}) \\ 961 = n^2 \\ Extracting~the~square~root~you'll~have: \\ n = +-31 \\ Take~the~positive~value~since~n~should~be~positive. \\ Therefore~n=31[/tex]

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