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ધારો કે $(1+x)^{10}$ ના વિસ્તરણમાં $x^{ r }$ નો દ્વિપદ્દી સહગગણક $C _{ r }$ વડે દર્શાવાય છે. જો $\alpha, \beta \in R$ માટે, $C _{1}+3 \cdot 2 C _{2}+5 \cdot 3 C _{3}+\ldots 10$ પદો સુધી = $\frac{\alpha \times 2^{11}}{2^{\beta}-1}\left(C_{0}+\frac{C_{1}}{2}+\frac{C_{2}}{3}+\ldots 10\right.$ પદો સુધી $)$, તો $\alpha+\beta$ ની કિમત ....... છે.
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BONUS PLUSE
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Solution
$(1+x)^{10}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots \ldots+C_{10} x^{10}$
Differentiating
$10(1+x)^{9}=C_{1}+2 C_{2} x+3 C_{3} x^{2}+\ldots+10 C_{10} x^{9}$
replace $x \rightarrow X ^{2}$
$10\left(1+x^{2}\right)^{9}=C_{1}+2 C_{2} x^{2}+3 C_{3} x^{4}+\ldots+10 C_{10} x^{18}$
$10 \cdot x\left(1+x^{2}\right)^{9}=C_{1} x+2 C_{2} x^{3}+3 C_{3} x^{5}+\ldots .+10 C_{10} x^{19}$
Differentiating
$10\left(\left(1+x^{2}\right)^{9} \cdot 1+x \cdot 9\left(1+x^{2}\right)^{8} 2 x\right)$
$=C_{1} x+2 C_{2} \cdot 3 x^{3}+3 \cdot 5 \cdot C_{3} x^{4}+\ldots .+10 \cdot 19 C_{10} x^{18}$
putting $x=1$
$10\left(2^{9}+18 \cdot 2^{8}\right)$
$= C _{1}+3 \cdot 2 \cdot C _{2}+5 \cdot 3 \cdot C _{3}+\ldots+19 \cdot 10 \cdot C _{10} $
$C _{1}+3 \cdot 2 \cdot C _{2}+\ldots \ldots+19 \cdot 10 \cdot C _{10}$
$=10 \cdot 2^{9} \cdot 10=100 \cdot 2^{9}$
$C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots . .+\frac{ C _{9}}{11}+\frac{ C _{10}}{11}=\frac{2^{11}-1}{11}$
$10^{\text {th }} \text { term } 11^{\text {th }} \text { term }$
$C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots .+\frac{ C _{9}}{11}=\frac{2^{11}-2}{11}$
Now, $100 \cdot 2^{9}=\frac{\alpha \cdot 2^{11}}{2^{\beta}-1}\left(\frac{2^{11}-2}{11}\right)$
Eqn. of form $y = k \left(2^{ x }-1\right)$.
It has infinite solutions even if we take $x, y \in N$.