\(∑_{r=1}^{26} \frac{1}{(2r−1)!(51−(2r−1))!} \)
\(= ∑_{r=1}^{26} ^{51}C_{(2r−1)} \frac{1}{51!}\)
\(=\frac{1}{51!} {^{51}C_1 + ^{51}C_3 +….+ ^{51}C_{51}}= \frac{1}{51!} (2^{50})\)
\(= (\frac{2^{50}}{51!} )\)
Therefore, The correct answer is option (D): \(\frac{2^{50}}{51!} \)
\[ \sum_{r=1}^{26} \frac{1}{(2r-1)! (51-(2r-1))!} = \sum_{r=1}^{26} \binom{51}{2r-1} \frac{1}{51!} \] \[ = \frac{1}{51!} \sum_{r=1}^{26} \binom{51}{2r-1} = \frac{1}{51!} (2^{50}) \]
A molecule with the formula $ \text{A} \text{X}_2 \text{Y}_2 $ has all it's elements from p-block. Element A is rarest, monotomic, non-radioactive from its group and has the lowest ionization energy value among X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
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Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.
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Definite integral is an operation on functions which approximates the sum of the values (of the function) weighted by the length (or measure) of the intervals for which the function takes that value.
Definite integrals - Important Formulae Handbook
A real valued function being evaluated (integrated) over the closed interval [a, b] is written as :
\(\int_{a}^{b}f(x)dx\)
Definite integrals have a lot of applications. Its main application is that it is used to find out the area under the curve of a function, as shown below: