Skip to content
dronacharya.ai
← JEE Main 2022 · 25 Jun · Shift 1
MathematicsInteger+4 / -1Question 21 of 90

JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 21

Let CrC_r denote the binomial coefficient of xrx^r in the expansion of (1+x)10(1+x)^{10}. If for α,β∈R\alpha,\beta\in\mathbb R,
C1+3⋅2 C2+5⋅3 C3+… upto 10 terms=α×2112β−1(C0+C12+C23+… upto 10 terms)C_1+3\cdot2\,C_2+5\cdot3\,C_3+\dots\text{ upto 10 terms}=\frac{\alpha\times2^{11}}{2^\beta-1}\left(C_0+\frac{C_1}2+\frac{C_2}3+\dots\text{ upto 10 terms}\right)
then the value of α+β\alpha+\beta is equal to ______.

Answer and worked solution

Official answer: 286

Left side. The rr-th term is r(2r−1)Crr(2r-1)C_r for r=1,…,10r=1,\dots,10: 2∑r2Cr−∑rCr=2(10⋅11⋅28)−10⋅29=512002\sum r^2C_r-\sum rC_r=2(10\cdot11\cdot2^8)-10\cdot2^9=51200.

Bracket. The key takes the bracket as the full sum ∑r=010Crr+1=211−111\displaystyle\sum_{r=0}^{10}\frac{C_r}{r+1}=\frac{2^{11}-1}{11} (11 terms; with only 10 terms no integer β\beta works).

Match. α 2112β−1⋅211−111=51200\dfrac{\alpha\,2^{11}}{2^\beta-1}\cdot\dfrac{2^{11}-1}{11}=51200 with β=11\beta=11 gives α=51200×112048=275\alpha=\dfrac{51200\times11}{2048}=275.

Sum. α+β=286\alpha+\beta=286.

Answer: 286

Topics:Properties of binomial coefficients
← Previous questionNext question →