Let $$C_r$$ denote the binomial coefficient of $$x^r$$ in the expansion of $$(1 + x)^{10}$$. If for $$\alpha, \beta \in R$$,
$$C_{1} + 3 \cdot 2C_{2} + 5 \cdot 3C_{3} + \ldots $$ upto 10 terms $$= \frac{\alpha \times 2^{11}}{2^{\beta} - 1}(C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \ldots $$ upto 10 terms ) then the value of $$\alpha + \beta $$ is equal to ______.

Given,

$$C_r=\binom{10}{r}$$

Let

$$S=C_{1}+3\cdot2C_{2}+5\cdot3C_{3}+\cdots$$ upto $$10$$ terms

Now,

$$(1+x)^{10}=C_0+C_1x+C_2x^2+\cdots+C_{10}x^{10}$$

Differentiating,

$$10(1+x)^9=C_1+2C_2x+3C_3x^2+\cdots+10C_{10}x^9$$

Replace $$x$$ by $$x^2,$$

$$10(1+x^2)^9=C_1+2C_2x^2+3C_3x^4+\cdots+10C_{10}x^{18}$$

Multiplying by $$x,$$

$$10x(1+x^2)^9=C_1x+2C_2x^3+3C_3x^5+\cdots+10C_{10}x^{19}$$

Differentiating,

$$10\left((1+x^2)^9+x\cdot9(1+x^2)^8\cdot2x\right)=C_1+3\cdot2C_2x^2+5\cdot3C_3x^4+\cdots+19\cdot10C_{10}x^{18}$$

Putting $$x=1,$$

$$10\left(2^9+18\cdot2^8\right)=C_1+3\cdot2C_2+5\cdot3C_3+\cdots+19\cdot10C_{10}$$

Hence,

$$S=10\cdot2^8(2+18)$$

$$=10\cdot20\cdot2^8$$

$$=100\cdot2^9$$

Now,

$$T=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\cdots+\frac{C_{10}}{11}$$

$$=\sum_{r=0}^{10}\frac{C_r}{r+1}$$

Since,

$$C_r=\binom{10}{r}$$

we use the identity

$$\frac1{r+1}\binom{10}{r}=\frac1{11}\binom{11}{r+1}$$

Therefore,

$$T=\frac1{11}\sum_{r=0}^{10}\binom{11}{r+1}$$

Now put

$$k=r+1$$

Then,

$$T=\frac1{11}\sum_{k=1}^{11}\binom{11}{k}$$

Using,

$$\sum_{k=0}^{11}\binom{11}{k}=2^{11}$$

we get

$$\sum_{k=1}^{11}\binom{11}{k}=2^{11}-1$$

Hence,

$$T=\frac{2^{11}-1}{11}$$

Previously,

$$S=C_1+3\cdot2C_2+5\cdot3C_3+\cdots+19\cdot10C_{10}=100\cdot2^9$$

Given,

$$S=\frac{\alpha\cdot2^{11}}{2^\beta-1}T$$

Substituting the values of $$S$$ and $$T,$$

$$100\cdot2^9=\frac{\alpha\cdot2^{11}}{2^\beta-1}\cdot\frac{2^{11}-1}{11}$$

Comparing denominator terms,

$$2^\beta-1=2^{11}-1$$

Hence,

$$\beta=11$$

Now,

$$100\cdot2^9=\frac{\alpha\cdot2^{11}}{11}$$

Dividing by $$2^9,$$

$$100=\frac{4\alpha}{11}$$

$$1100=4\alpha$$

$$\alpha=275$$

Therefore,

$$\alpha+\beta=275+11=286$$

Hence, $$\boxed{286}$$.