Let $(x+10)^{50}+(x-10)^{50}=\mathrm{a}_{0}+\mathrm{a}_{1} x+\mathrm{a}_{2} x^{2}+\ldots .+\mathrm{a}_{50} x^{50}$, for all $x \in \mathbf{R} ;$ then $\frac{\mathrm{a}_{2}}{\mathrm{a}_{0}}$ is equal to :

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Aicharya · Accepted Answer

$(x+10)^{50}+(x-10)^{50}$ $=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{50} x^{50}$ $	herefore \quad a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{50} x^{50}$ $=2\left({ }^{50} C_{0} x^{50}+{ }^{50} C_{2} x^{48} \cdot 10^{2}+{ }^{50} C_{4} x^{46} \cdot 10^{4}+\ldotsight)$ $	herefore \quad a_{0}=2 \cdot{ }^{50} C_{50} 10^{50}$ $\quad a_{2}=2 \cdot{ }^{50} C_{2} \cdot 10^{48}$ $	herefore \quad \frac{a_{2}}{a_{0}}=\frac{{ }^{50} C_{2} 	imes 10^{48}}{{ }^{50} C_{50} 10^{50}}$ $=\frac{50 	imes 49}{2 	imes 100}=\frac{49}{4}=12.25$

Let $(x+10)^{50}+(x-10)^{50}=\mathrm{a}_{0}+\mathrm{a}_{1} x+\mathrm{a}_{2} x^{2}+\ldots .+\mathrm{a}_{50}…

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