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A quiz on binomial therorm +1 ncert JEE level

All 25 questions from this finished quiz, with the correct answer and explanation. Finished 16 Oct 2026, 3:34 am IST.
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  1. 1. What is the coefficient of x50x^{50} in (1+x)1000+2x(1+x)999+3x2(1+x)998+⋯+1001x1000(1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998}+\dots+1001x^{1000}?

    A. (100250)\binom{1002}{50}

    B. (100150)\binom{1001}{50}

    C. (100050)\binom{1000}{50}

    D. (100251)\binom{1002}{51}

    Answer: A. (100250)\binom{1002}{50}

    Step-by-step solution

    1. 1The term (k+1)xk(1+x)1000−k(k+1)x^k(1+x)^{1000-k} contributes (k+1)(1000−k950)(k+1)\binom{1000-k}{950}, so the coefficient is ∑k=050(k+1)(1000−k950)\sum_{k=0}^{50}(k+1)\binom{1000-k}{950}
    2. 2Write k+1=∑j=0k1k+1=\sum_{j=0}^{k}1 and swap the sums: ∑j=050∑k=j50(1000−k950)=∑j=050(1001−j951)\sum_{j=0}^{50}\sum_{k=j}^{50}\binom{1000-k}{950}=\sum_{j=0}^{50}\binom{1001-j}{951} (hockey-stick)
    3. 3Applying hockey-stick again: =(1002952)=(100250)=\binom{1002}{952}=\binom{1002}{50}
  2. 2. If Cr=(nr)C_r=\binom{n}{r}, what is the value of C0+C12+C23+⋯+Cnn+1C_0+\frac{C_1}{2}+\frac{C_2}{3}+\dots+\frac{C_n}{n+1}?

    A. 2n+1−1n+1\frac{2^{n+1}-1}{n+1}

    B. 2n+1n+1\frac{2^{n+1}}{n+1}

    C. 2n−1n+1\frac{2^{n}-1}{n+1}

    D. 2n+1−1n\frac{2^{n+1}-1}{n}

    Answer: A. 2n+1−1n+1\frac{2^{n+1}-1}{n+1}

    Step-by-step solution

    1. 11r+1(nr)=1n+1(n+1r+1)\frac{1}{r+1}\binom{n}{r}=\frac{1}{n+1}\binom{n+1}{r+1}
    2. 2Sum =1n+1[(n+11)+(n+12)+⋯+(n+1n+1)]=\frac{1}{n+1}\left[\binom{n+1}{1}+\binom{n+1}{2}+\dots+\binom{n+1}{n+1}\right]
    3. 3=2n+1−1n+1=\frac{2^{n+1}-1}{n+1}

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