Question

The value of $\underset{n\to \infty }{lim}\underset{k=1}{\sum ^n}(\frac{n-k}{n^2})\cos \frac{4k}{n}=\frac{1}{a}(1-\cos c),$ then

• A. $a+c=20$
• B. $\frac{a}{c}=4$
• C. $a-c=12$
• D. $ac=16$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a+c=20$
B $\frac{a}{c}=4$
C $a-c=12$

$S=\frac{1}{n}\underset{k=1}{\sum ^n}(1-\frac{k}{n})\cos 4\left(\frac{k}{n}\right)={\int }_0^1(1-x)\cos 4xdx$
$=[(1-x)\frac{\sin 4x}{4}{]}_0^1+\frac{1}{4}{\int }_0^1\sin 4xdx$
$=(\frac{-1}{16}\cos 4x{)}_0^1=\frac{1}{16}(1-\cos 4)$