Step 1: We are given a sum involving factorial terms, and we need to compute the limit as \( n \to \infty \). We can simplify the expression by examining the asymptotic behavior of the sum.
Step 2: For large \( k \), the term \( (k + 3)! \) grows very quickly compared to the polynomial terms in the numerator. Thus, the terms of the sum decrease rapidly as \( k \) increases.
Step 3: We recognize that the sum is dominated by the first few terms, and we compute the value of the infinite sum by summing the first few terms and taking the limit. The value of the sum as \( n \to \infty \) is \( \frac{4}{3} \). Thus, the correct answer is (1).
\[ 5m \sum_{r=m}^{2m} T_r \text{ is equal to:} \]
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to:
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,