The value of $\frac{1 \times 2^2 + 2 \times 3^2 + \dots + 100 \times (101)^2}{1^2 \times 2 + 2^2 \times 3 + \dots + 100^2 \times 101}$ is:
We are asked to evaluate the expression:
\[ \frac{1 \times 2^2 + 2 \times 3^2 + \cdots + 100 \times (101)^2}{1^2 \times 2^2 + 2^2 \times 3^2 + \cdots + 100^2 \times 101}. \]
This can be rewritten as:
\[ \frac{\sum_{r=1}^{100} r(r+1)^2}{\sum_{r=1}^{100} r^2(r+1)}. \]
Now, expand both the numerator and denominator:
Numerator: \[ \sum_{r=1}^{100} r(r+1)^2 = \sum_{r=1}^{100} r(r^2 + 2r + 1) = \sum_{r=1}^{100} (r^3 + 2r^2 + r). \] Denominator: \[ \sum_{r=1}^{100} r^2(r+1) = \sum_{r=1}^{100} (r^3 + r^2). \]
We now need to compute these sums:
\[ \sum_{r=1}^{100} r^3 = \left(\frac{100(100+1)}{2}\right)^2 = 25502500. \] \[ \sum_{r=1}^{100} r^2 = \frac{100(100+1)(2 \times 100+1)}{6} = 338350. \]
Using these values, we can calculate:
Numerator: \[ 25502500 + 2 \times 338350 + 5050 = 51851000. \] Denominator: \[ 25502500 + 338350 = 25840850. \]
Thus, the value of the expression is:
\[ \frac{51851000}{25840850} = \frac{305}{301}. \]
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,