So, The correct option is (B): $\frac7{29}$
\(\overset{21}{\underset{n=1}\sum} \frac3{(4n-1)(4n+3)} = \frac3{4}\overset{21}{\underset{n=1}\sum} \frac1{(4n-1)}-\frac1{(4n+3)}\)
=43n=1∑21(4n−1)(4n+3)(4n+3)−(4n−1)
=43n=1∑214n−11−4n+31
=43(31−71+71−111+111−….+831−871)
=43(31−871)=297
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32
A collection of numbers that is presented as the sum of the numbers in a stated order is called a series. As an outcome, every two numbers in a series are separated by the addition (+) sign. The order of the elements in the series really doesn't matters. If a series demonstrates a finite sequence, it is said to be finite, and if it demonstrates an endless sequence, it is said to be infinite.
Read More: Sequence and Series
The following are the two main types of series are: