माना $C _{ r },(1+ x )^{10}$ के प्रसार में $x ^{ r }$ के द्विपद गुणांक को प्रदर्शित करता है। यदि $\alpha, \beta \in R$ के लिए

$C _1+3.2 C _2+5 \cdot 3 C _3+\ldots 10$ पद तक

$=\frac{\alpha \times 2^{11}}{2^\beta-1}( C _0+\frac{ C _1}{2}+\frac{ C _2}{3}+\ldots . .10$ पद तक है,तो $\alpha+\beta$ का मान होगा

$(1+x)^{10}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots \ldots+C_{10} x^{10}$

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 $x, y \in N$.