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Sum of (a polynomial in r) times nCr, using differentiation

This type of question, asked 8 times: 2021 Paper 10, 2021 Paper 18, 2021 Paper 25, 24 Jan 2023, 8 Apr 2024, 3 Apr 2025, 4 Apr 2025, 7 Apr 2025. All 8 questions below, oldest first.

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  1. Question 1

    2021 Paper 10

    88
    Let  nCr\,^{n}C_{r} denote the binomial coefficient of xrx^{r} in the expansion of (1+x)n(1 + x)^{n}. If ∑k=010(22+3k) 10Ck=α⋅310+β⋅210, α,β∈R\sum_{k = 0}^{10} \left( 2^{2}+ 3k \right)\,^{10}C_{k} = \alpha\cdot 3^{10}+\beta\cdot 2^{10},\ \alpha,\beta\in R, then α+β\alpha+\beta is equal to

  2. Question 2

    2021 Paper 18

    69
    If  20Cr\ ^{20}C_{r} is the co-efficient of xrx^{r} in the expansion of (1+x)20(1 +x)^{20}, then the value of ∑r=020r2 20Cr\sum_{r= 0}^{20} r^{2}\ ^{20}C_{r} is equal to :

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  3. Question 3

    2021 Paper 25

    83
    Let m,n∈Nm,n\in N and gcd(2,n)=1gcd(2,n) = 1. If 30(300)+29(301)+…+2(3028)+1(3029)=n⋅2m30\begin{pmatrix} 30 \\ 0 \end{pmatrix}+ 29\begin{pmatrix} 30 \\ 1 \end{pmatrix}+ \ldots + 2\begin{pmatrix} 30 \\ 28 \end{pmatrix}+ 1\begin{pmatrix} 30 \\ 29 \end{pmatrix}=n\cdot2^{m}, then n+mn+m is equal to

  4. Question 4

    24 Jan 2023

    84
    Suppose ∑r=02023r2 2023Cr=2023×α×22022\sum_{r = 0}^{2023} r^{2}\ ^{2023}C_{r}= 2023 \times \alpha \times2^{2022}. Then the value of α\alpha is

  5. Question 5

    8 Apr 2024

    87
    Let α=∑r=0n(4r2+2r+1) nCr\alpha =\sum_{r = 0}^{n} \left( 4r^{2}+ 2r + 1 \right)\ ^{n}C_{r} and β=(∑r=0n nCrr+1)+1n+1\beta =\left( \sum_{r = 0}^{n} \frac{\ ^{n}C_{r}}{r + 1} \right)+\frac{1}{n + 1}. If 140<2αβ<281140 <\frac{2\alpha}{\beta}< 281, then the value of nn is

  6. Question 6

    3 Apr 2025

    59
    If ∑r=19(r+32r)⋅ 9Cr=α(32)9−β, α,β∈N\sum_{r = 1}^{9} \left( \frac{r + 3}{2^{r}} \right)\cdot\ ^{9}C_{r}=\alpha\left( \frac{3}{2} \right)^{9}-\beta,\ \alpha,\beta\in N, then (α+β)2(\alpha+\beta)^{2} is equal to.

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  7. Question 7

    4 Apr 2025

    138
    If 12⋅( 15C1)+22⋅( 15C2)+32⋅( 15C3)1^{2}\cdot\left( \ ^{15}C_{1} \right)+2^{2}\cdot\left( \ ^{15}C_{2} \right)+3^{2}\cdot\left( \ ^{15}C_{3} \right) +….+152⋅( 15C15)=+ \ldots. +15^{2}\cdot\left( \ ^{15}C_{15} \right)= 2m⋅3n⋅5k2^{m}\cdot3^{n}\cdot5^{k}, where m,n,k∈Nm,n,k \in N, then m+n+km + n + k is equal to :-

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  8. Question 8

    7 Apr 2025

    150
    The sum of the series
    2×1× 20C4−3×2× 20C5+4×3× 20C6−5×4× 20C7+…+18×17× 20C20, is equal to .\begin{matrix} & 2 \times 1 \times \ ^{20}C_{4} - 3 \times 2 \times \ ^{20}C_{5} + 4 \times 3 \times \ ^{20}C_{6} - 5 \times 4 \times \\ & \ ^{20}C_{7} + \ldots + 18 \times 17 \times \ ^{20}C_{20},\text{~is equal to~}\text{.} \end{matrix}