Mathematics · 2019
JEE Main · 11 January 2019, Shift 2 · Q69
Let S_n =1+ q + q^2+…+ q^n and T_n =1+((q +1)/2)+((q +1)/2)^2+…+((q +1)/2)^n where q is a real number and q ≠ 1. If ^101 C_1+^101 C_2 · S_1+…+^101…
Let $\displaystyle \mathrm{S}_{\mathrm{n}}=1+\mathrm{q}+\mathrm{q}^2+\ldots+\mathrm{q}^{\mathrm{n}}$ and $\displaystyle \mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{q}+1}{2}\right)+\left(\frac{\mathrm{q}+1}{2}\right)^2+\ldots+\left(\frac{\mathrm{q}+1}{2}\right)^{\mathrm{n}}$
where q is a real number and $\displaystyle \mathrm{q} \neq 1$. If $\displaystyle { }^{101} \mathrm{C}_1+{ }^{101} \mathrm{C}_2 \cdot \mathrm{~S}_1+\ldots+{ }^{101} \mathrm{C}_{101} \cdot \mathrm{~S}_{100}=\alpha \mathrm{T}_{100}$
then $\displaystyle \alpha$ is equal to :
Official answer
(3)
$\displaystyle 2^{100}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.