If $\left({ }^{30} C _1\right)^2+2\left({ }^{30} C _2\right)^2+3\left({ }^{30} C _3\right)^2+\ldots+30\left({ }^{30} C _{30}\right)^2=\frac{\alpha 60 !}{(30 !)^2}$ then $\alpha$ is equal to :
- 60
- 10
- 15
- 30
The Correct Option is C
Approach Solution - 1
Approach Solution -2
Step 1: Expand the series
The given series is:
\[ S = \binom{30}{0} + 2 \cdot \binom{30}{1} \cdot 2 + 3 \cdot \binom{30}{2} \cdot 2 + \dots + 30 \cdot \binom{30}{30} \cdot 2. \]
Each term can be written as:
\[ n \left( \binom{30}{n} \right) 2. \]
Step 2: Use the identity for weighted sums
The identity for such sums is:
\[ \sum_{k=0}^{n} k \cdot \left( \binom{n}{k} \right)^2 = n \cdot \binom{2n-1}{n-1}. \]
Substitute \( n = 30 \):
\[ S = 30 \cdot \binom{59}{29}. \]
Step 3: Express \( \binom{59}{29} \) in factorials
Using the formula for combinations:
\[ \binom{59}{29} = \frac{59!}{29! \cdot 30!}. \]
Thus:
\[ S = 30 \cdot \frac{59!}{29! \cdot 30!}. \]
Step 4: Compare with the given expression
The series is given as:
\[ S = \alpha \cdot \frac{60!}{(30!)^2}. \]
Substitute \( 60! = 60 \cdot 59! \):
\[ S = \alpha \cdot \frac{60 \cdot 59!}{(30!)^2}. \]
Equating the two expressions \[ 30 \cdot \frac{59!}{29! \cdot 30!} = \alpha \cdot \frac{60 \cdot 59!}{(30!)^2}. \] Simplify: \[ 30 \cdot 29! \cdot 30 = \alpha \cdot 60. \] \[ \alpha = \frac{30 \cdot 30}{60} = 15. \]
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Concepts Used:
Oxidation Number
Oxidation number, also called oxidation state, the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.
Oxidation number of an atom is defined as the charge that an atom appears to have on forming ionic bonds with other heteroatoms. An atom having higher electronegativity (even if it forms a covalent bond) is given a negative oxidation state.
The definition, assigns oxidation state to an atom on conditions, that the atom –
- Bonds with heteroatoms.
- Always form ionic bonding by either gaining or losing electrons, irrespective of the actual nature of bonding.
Oxidation number is a formalized way of keeping track of oxidation state.
Read More: Oxidation and Reduction
Way To Find Oxidation Number Of An Atom?
Oxidation number or state of an atom/ion is the number of electrons an atom/ion that the molecule has either gained or lost compared to the neutral atom. Electropositive metal atoms, of group I, 2 and 3 lose a specific number of electrons and have always constant positive oxidation numbers.
In molecules, more electronegative atom gain electrons from a less electronegative atom and have negative oxidation states. The numerical value of the oxidation state is equal to the number of electrons lost or gained.
Oxidation number or oxidation state of an atom or ion in a molecule/ion is assigned by:
- Summing up the constant oxidation state of other atoms/molecules/ions that are bonded to it and
- Equating, the total oxidation state of a molecule or ion to the total charge of the molecule or ion.