Question:
The average of the coefficient of the two middle terms in the expansion of \((1+x)^{2n+1}\) is:
\((1+x)^{2n+1}\) के प्रसार में दो मध्य पदों के गुणांकों का औसत है:
Options:
- \({}^{2n+1}C_{n+2}\)
- \({}^{2n+1}C_n\)
- \({}^{2n+1}C_{n-1}\)
- \({}^{2n}C_{n+1}\)
- \({}^{2n+1}C_{n+2}\)
- \({}^{2n+1}C_n\)
- \({}^{2n+1}C_{n-1}\)
- \({}^{2n}C_{n+1}\)
Correct Answer: b)
Explanation:
In \((1+x)^{2n+1}\), the power \(2n+1\) is odd, so there are \((2n+1)+1 = 2n+2\) terms (an even number of terms).
The two middle terms are the \((n+1)^{\text{th}}\) and \((n+2)^{\text{th}}\) terms.
Their coefficients are \({}^{2n+1}C_n\) and \({}^{2n+1}C_{n+1}\).
By the symmetry property \({}^N C_R = {}^N C_{N-R}\):
\[{}^{2n+1}C_{n+1} = {}^{2n+1}C_{(2n+1)-(n+1)} = {}^{2n+1}C_n\]
Therefore, their average is:
\[\frac{{}^{2n+1}C_n + {}^{2n+1}C_{n+1}}{2} = \frac{{}^{2n+1}C_n + {}^{2n+1}C_n}{2} = {}^{2n+1}C_n\]
Hence, the correct option is (b).
\((1+x)^{2n+1}\) में कुल पदों की संख्या \(2n+2\) (सम) है।
अतः दो मध्य पद \((n+1)\)वाँ और \((n+2)\)वाँ पद हैं, जिनके गुणांक क्रमशः \({}^{2n+1}C_n\) तथा \({}^{2n+1}C_{n+1}\) हैं।
सममिति गुणधर्म से \({}^{2n+1}C_{n+1} = {}^{2n+1}C_n\)।
अतः दोनों का औसत:
\[\frac{{}^{2n+1}C_n + {}^{2n+1}C_n}{2} = {}^{2n+1}C_n\]
अतः सही विकल्प (b) है।