JEE Main Algebra — Combinatorics & Series: Binomial Theorem and its Applications – Chapter Test

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Difficulty: 10 medium · 10 hard

Binomial Theorem and its Applications MCQ questions with answers

The first 3 of 20 questions. Unlock the series for the rest, each with a solution.

1. Q1. In the expansion of (1 + x)^15, the greatest binomial coefficient occurs in:

   (1 + x)^15 के प्रसार में, सबसे बड़ा द्विपद गुणांक किसमें आता है:

   • A. the 16th term / 16वें पद में
   • B. the 8th term only / केवल 8वें पद में
   • C. the 7th and 8th terms / 7वें और 8वें पदों में
   • D. the 8th and 9th terms / 8वें और 9वें पदों में
   Solution

   Since 15 is odd, the greatest binomial coefficients are C(15,7) = C(15,8), which are the coefficients of the 8th and 9th terms.

   चूँकि 15 विषम है, सबसे बड़े द्विपद गुणांक C(15,7) = C(15,8) हैं, जो 8वें और 9वें पदों के गुणांक हैं।

2. Q2. The number of distinct terms in the expansion of (a + b + c)^10 is:

   (a + b + c)^10 के प्रसार में भिन्न-भिन्न पदों की संख्या है:

   • A. 33
   • B. 11
   • C. 66
   • D. 55
   Solution

   The number of terms in (x1 + ... + xk)^n is C(n+k−1, k−1); here C(10+2, 2) = C(12,2) = 66.

   (x1 + ... + xk)^n में पदों की संख्या C(n+k−1, k−1) होती है; यहाँ C(10+2, 2) = C(12,2) = 66।

3. Q3. If (1 + x)^n = C0 + C1x + C2x² + ... + Cnx^n and C1 + C2 + ... + Cn = 255, then the value of n is:

   यदि (1 + x)^n = C0 + C1x + C2x² + ... + Cnx^n और C1 + C2 + ... + Cn = 255, तो n का मान है:

   • A. 8
   • B. 9
   • C. 10
   • D. 7
   Solution

   The sum of all binomial coefficients is 2^n, so C1 + C2 + ... + Cn = 2^n − 1 = 255, giving 2^n = 256 and n = 8.

   सभी द्विपद गुणांकों का योग 2^n होता है, अतः C1 + C2 + ... + Cn = 2^n − 1 = 255, जिससे 2^n = 256 और n = 8 प्राप्त होता है।

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