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.

  • [JEE MAIN 2022]
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$(1+x)^{10}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots \ldots+C_{10} x^{10}$

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 $x, y \in N$.

Similar Questions

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$(1-x)^{100}$ ના દ્વિપદી વિસ્તરણમાં પ્રથમ $50$ પદોના સહગુણકોનો સરવાળો $.......$ છે.

  • [JEE MAIN 2023]

$\frac{{{C_0}}}{1} + \frac{{{C_1}}}{2} + \frac{{{C_2}}}{3} + .... + \frac{{{C_n}}}{{n + 1}} = $

જો $(1 -x + x^2)^n = a_0 + a_1x + a_2x^2 + ....... + a_{2n}x^{2n}$,હોય તો  $a_0 + a_2 + a_4 +........+ a_{2n}$ ની કિમત મેળવો 

જો ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .......... + {C_n}{x^n}$, તો $\frac{{{C_1}}}{{{C_0}}} + \frac{{2{C_2}}}{{{C_1}}} + \frac{{3{C_3}}}{{{C_2}}} + .... + \frac{{n{C_n}}}{{{C_{n - 1}}}} = $