If \(\mathrm{C}_{\mathrm{j}}\) stands for \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{j}}\), then
\(\frac{\mathrm{C}_0}{2}+\frac{\mathrm{C}_1}{2.2^2}+\frac{\mathrm{C}_2}{3.2^3}+\ldots+\frac{\mathrm{C}_{\mathrm{n}}}{(\mathrm{n}+1) 2^{\mathrm{n}+1}}=\)

The number of ways in which \(n\) distinct objects can be put into two different boxes is

If \(\int \frac{1}{1-\cos x} d x=\tan \left(\frac{x}{\alpha}+\beta\right)+c\), then one of the values of \(\frac{\pi \alpha}{4}-\beta\) is

If \(2 \alpha=-1-i \sqrt{3}\) and \(2 \beta=-1+i \sqrt{3}\), then \(5 \alpha^4+5 \beta^4+7 \alpha^{-1} \beta^{-1}\) is equal to

The harmonic conjugate of the point \((2,3,4)\) with respect to the point \((3,-2,2)\) and \((6,-17,-4)\) is

If \(\theta=\operatorname{Tan}^{-1}\left(\frac{1}{3}\right)+\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)+\operatorname{Tan}^{-1}\left(\frac{1}{13}\right)+\operatorname{Tan}^{-1}\left(\frac{1}{21}\right)+\operatorname{Tan}^{-1}\left(\frac{1}{31}\right)\), then \(\tan \theta=\)

Let \(\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\)
Assertion (A) : The identity
\(|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}|^2+|\overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}|^2=2|\overrightarrow{\mathbf{a}}|^2\) holds for
\(\vec{a}\).
Reason (R) : \(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{i}}=a_3 \hat{\mathbf{j}}-a_2 \hat{\mathbf{k}}\),
\(\overrightarrow{\mathbf{a}} \times \hat{\mathbf{j}}=a_1 \hat{\mathbf{k}}-a_3 \hat{\mathbf{i}}, \overrightarrow{\mathbf{a}} \times \hat{\mathbf{k}}=a_2 \hat{\mathbf{i}}-a_1 \hat{\mathbf{j}}\)
Which of the following is correct?

\(\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k}\left({ }^k C_r\right)\) is equal to

If \(\mathrm{O}(0,0,0), \mathrm{A}(3,0,0), \mathrm{B}(0,4,0)\) form a triangle then the incentre of triangle OAB is

Match the following
\(\begin{array}{llll}
& \text {List - I} & & \text {List - II} \\
\text {a)} & \text {See Saw Shape} & \text {i)} & \mathrm{XeF}_4 \\
\text {b)} & \text {Square Pyramidal} & \text {ii)} & \mathrm{CF}_3 \\
\text {c)} & \text {T-Shape} & \text {iii)} & \mathrm{PbCl}_2 \\
\text {d)} & \text{Bent Shape} & \text {iv)} & \mathrm{SF}_4 \\
& & \text {v)} & \mathrm{BrF}_5
\end{array}\)
The correct answer is

In the expansion of \((\sqrt[5]{3}+\sqrt[3]{2})^{15}\)

If the equation to the locus of points equidistant from the points \((-2,3),(6,-5)\) is \(a x+b y+c=0\), where \(a>0\), then the ascending order of \(a, b, c\) is

Which of the following is not correct?