If $$1^2+2^2+3^2+………+14^2=1015,$$​ then $$3^2+6^2+9^2+………..+42^2$$​ is equal to:

Correct option is A

Given:
​$$1^2+2^2+3^2+………+14^2=1015,$$​​
Solution:
series:
​$$1^2+2^2+3^2+………+14^2=1015,$$
Now, let's look at the second series:
​$$3^2+6^2+9^2+………..+42^2$$​​
We can factor out 3² from each term:
3² $$\times$$​ (1² + 2² + 3² + ... + 14²)
Now, we can substitute the given value of the first series:
= 9 $$\times$$​ 1015
= 9135