Question

# Find the sum of $$n$$ terms of the series whose $$nth$$ term is :$$(2n-1)^{2}$$

Solution

## $$\sum\ (2n-1)^2$$$$\sum 4n^2-4n+1$$$$4\sum n^2-4\sum\ n+\sum1$$$$\dfrac{4n({n+1})(2n+1)}{6}+\dfrac{4n(n+1)}{2}+n$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More