Answer:
The sum of the sequence 1+3+5+7+...1001 is 251001.
Step-by-step explanation:
Given sequence is 1 + 3 + 5 + 7 + ....+ 1001.
Here, \(a_0=1,a_1=3,a_2=5\) and so on upto \(a_n=1001\)
Lets find the difference between each term
\(a_1-a_0=3-1=2\)
\(a_3-a_2=5-3=2\)
\(a_4-a_3=7-5=2\)
We see that the difference between each term of the given sequence is 2 . Thus, it is an Arithmetic sequence.
Since we have to find the sum of the sequence
Sum of sequence of a given Arithmetic sequence is given as :
\(S_n=(a_0+a_n)\times \frac{n}{2}\) .............(1)
But, first find the number of terms,
\(a_n=a_0+(n-1)d\)
Put values, we get,
\(\Rightarrow 1001=1+(n-1)2\)
\(\Rightarrow 1001=1+2n-2\)
\(\Rightarrow 1001=2n-1\)
\(\Rightarrow 1001+1=2n\)
\(\Rightarrow 501=n\)
Now put values, \(a_n=1001\) , \(501=n\) and \(a_0=1\)
in (1), we get,
\(S_n=(a_0+a_n)\times \frac{n}{2}\)
\(\Rightarrow S_n=(1+1001)\times \frac{501}{2}\)
\(\Rightarrow S_n=(1002)\times \frac{501}{2}\)
\(\Rightarrow S_n=251001\)
Thus, the sum of the sequence 1+3+5+7+...1001 is 251001.
