Question

Series-2 sum =
1
/
2
+
4/

5
+
7/

8
+
10/

11 ……. 10 terms

Answer 1

Answer:

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Explanation:

If n is odd, then n−1 is even and we know that the sum of the first n−1 entries is 3(n−1)24. Moreover, the last entry in the sequence was the ((n−1)/2+1)-st entry of the first arithmetic progression, that is 1,4,7,10,…. But the value of that entry is 3((n−1)/2+1)−2=3(n−1)/2+1, so using the formula for the even case, the sum of the first n entries equals

3(n−1)24+3(n−1)/2+1=34(n2−2n+2n+1−2+4/3)=3n24+14.