Question

$1+5+9+\cdots \cdots +(4n-3)$ is equal to

• A. $n(4n-3)$
• B. $(2n-1)$
• C. $n(2n-1)$
• D. $(4n-3{)}^2$

Answer 1

The correct option is C $n(2n-1)$

In the given Arithmetic Progression,
First term $=a=1$
Common difference $=5-1=4$
Let $4n-3$ be the $k$ th term.
Then $x_n=a+(n-1)d$
$=>4n-3=1+(k-1)4$
$=>4n-1=1+4k-4$
$=>k=n$
So, $4n-3$ is the $n$ th term.
Now, Sum to 'n' terms of an AP $=\frac{n}{2}(2a+(n-1\left)d\right)=\frac{n}{2}(2+(n-1\left)4\right)=\frac{n}{2}(4n-2)=n(2n-1)$