how many terms of the sequence 1, 3, 5, 7, …. will give a sum of 961?

The number of terms is 31.

Step-by-step reasoning

You’re given the arithmetic sequence:

• 1,3,5,7,…1,3,5,7,\dots 1,3,5,7,…
• First term a=1a=1a=1
• Common difference d=2d=2d=2

You want the number of terms nnn such that the sum is 961. For an arithmetic sequence, the sum of the first nnn terms is:

Sn=n2 [2a+(n−1)d]S_n=\frac{n}{2},[2a+(n-1)d]Sn​=2n​[2a+(n−1)d]

Substitute a=1a=1a=1, d=2d=2d=2, and Sn=961S_n=961Sn​=961:

961=n2 [2⋅1+(n−1)⋅2]961=\frac{n}{2},[2\cdot 1+(n-1)\cdot 2]961=2n​[2⋅1+(n−1)⋅2]

961=n2 [2+2n−2]961=\frac{n}{2},[2+2n-2]961=2n​[2+2n−2]

961=n2 [2n]961=\frac{n}{2},[2n]961=2n​[2n]

961=n2961=n^2961=n2

So:

n2=961⇒n=31n^2=961\Rightarrow n=31n2=961⇒n=31

(We take the positive root because the number of terms must be positive.) So, 31 terms of the sequence 1,3,5,7,…1,3,5,7,\dots 1,3,5,7,… give a sum of 961.