Question

1+4+7+...+(3n-2)=n(3n-1)/2​

Answer 1

Answer:

Step-by-step explanation:

Sum of the first and last terms = 1+(3n−2)=3n−1

Sum of 2nd and (n-1)th terms = 4+(3n−5)=3n−1

Sum of 3rd and (n-2)th terms = 7+(3n−8)=3n−1

...

Sum of (n-1)th and 2nd terms = (3n−5)+4=3n−1

Sum of n-th (last) and 1st terms = (3n−2)+1=3n−1

Add both sides up .  

(1+4+...+(3n−2))+(1+4+...+(3n−2))=n(3n−1)

which means:

2(1+4+...+(3n−2))=n(3n−1)

or

1+4+...+(3n−2)=n(3n−1)/2

Answer 2

Answer: Check the image below for your answer.