1. Understanding Key Binomial Identities and the Problem Statement

This problem requires a strong understanding of binomial coefficient properties and summation techniques. We are given an equation involving sums of binomial coefficients $C_r$, where $C_r$ denotes ${}^{10}C_r$, the coefficient of $x^r$ in the expansion of ${(1 + x)^{10}}$. Our goal is to evaluate specific sums and then solve for $\alpha$ and $\beta$.

The fundamental properties of binomial coefficients and their sums that will be utilized are:

• Binomial Expansion: $(1+x)^n = \sum_{r=0}^n {}^n C_r x^r$
• Identity 1 (for $r \cdot C_r$): $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$
• Identity 2 (for $C_r / (r+1)$): $\frac{{}^n C_r}{r+1} = \frac{{}^{n+1} C_{r+1}}{n+1}$
• Sum of $r \cdot {}^n C_r$: $\sum_{r=1}^n r \cdot {}^n C_r = n \cdot 2^{n-1}$
• Sum of $r^2 \cdot {}^n C_r$: $\sum_{r=1}^n r^2 \cdot {}^n C_r = n(n-1)2^{n-2} + n2^{n-1}$
• Sum of $\frac{{}^n C_r}{r+1}$: $\sum_{r=0}^n \frac{{}^n C_r}{r+1} = \frac{2^{n+1}-1}{n+1}$

2. Evaluating the Left Hand Side (LHS) Expression

The given LHS expression is: $LHS = {C_1} + 3 \cdot 2{C_2} + 5 \cdot 3{C_3} + \dots \text{ upto 10 terms}$ Here, $C_r = {}^{10}C_r$. Let's observe the pattern of the terms. The general term can be written as $(2r-1) \cdot r \cdot {}^{10}C_r$. For $r=1$: $(2 \cdot 1 - 1) \cdot 1 \cdot C_1 = 1 \cdot C_1 = C_1$ For $r=2$: $(2 \cdot 2 - 1) \cdot 2 \cdot C_2 = 3 \cdot 2 \cdot C_2$ For $r=3$: $(2 \cdot 3 - 1) \cdot 3 \cdot C_3 = 5 \cdot 3 \cdot C_3$ This pattern holds true. Since there are 10 terms, $r$ ranges from $1$ to $10$.

So, we can write the LHS as a summation: $LHS = \sum_{r=1}^{10} (2r-1) r {}^{10}C_r$ Expand the term $(2r-1)r$: $LHS = \sum_{r=1}^{10} (2r^2 - r) {}^{10}C_r$ We can split this into two separate summations: $LHS = 2 \sum_{r=1}^{10} r^2 {}^{10}C_r - \sum_{r=1}^{10} r {}^{10}C_r$

Let's evaluate each sum separately using the identities. Here, $n=10$.

Derivation and Calculation of $\sum_{r=1}^{n} r \cdot {}^n C_r$ Consider the binomial expansion of $(1+x)^n$: $(1+x)^n = \sum_{r=0}^n {}^n C_r x^r = {}^n C_0 + {}^n C_1 x + {}^n C_2 x^2 + \dots + {}^n C_n x^n$ Differentiating both sides with respect to $x$: $n(1+x)^{n-1} = \sum_{r=1}^n r \cdot {}^n C_r x^{r-1}$ Now, substitute $x=1$ into this equation: $n(1+1)^{n-1} = \sum_{r=1}^n r \cdot {}^n C_r (1)^{r-1}$ $\sum_{r=1}^n r \cdot {}^n C_r = n \cdot 2^{n-1}$ For $n=10$: $\sum_{r=1}^{10} r \cdot {}^{10}C_r = 10 \cdot 2^{10-1} = 10 \cdot 2^9$

Derivation and Calculation of $\sum_{r=1}^{n} r^2 \cdot {}^n C_r$ We use the identity $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$. $\sum_{r=1}^n r^2 \cdot {}^n C_r = \sum_{r=1}^n r (r \cdot {}^n C_r)$ Substitute the identity: $= \sum_{r=1}^n r (n \cdot {}^{n-1} C_{r-1}) = n \sum_{r=1}^n r \cdot {}^{n-1} C_{r-1}$ To simplify the sum, let $k = r-1$. Then $r = k+1$. When $r=1, k=0$. When $r=n, k=n-1$. $= n \sum_{k=0}^{n-1} (k+1) {}^{n-1} C_k$ Split the sum: $= n \left( \sum_{k=0}^{n-1} k \cdot {}^{n-1} C_k + \sum_{k=0}^{n-1} {}^{n-1} C_k \right)$ The first sum is $\sum_{k=1}^{n-1} k \cdot {}^{n-1} C_k$, which is of the form $N \cdot 2^{N-1}$ with $N=n-1$. So, $\sum_{k=1}^{n-1} k \cdot {}^{n-1} C_k = (n-1)2^{(n-1)-1} = (n-1)2^{n-2}$. The second sum is the total sum of binomial coefficients for $n-1$: $\sum_{k=0}^{n-1} {}^{n-1} C_k = 2^{n-1}$. Thus, $\sum_{r=1}^n r^2 \cdot {}^n C_r = n \left( (n-1)2^{n-2} + 2^{n-1} \right)$ For $n=10$: $\sum_{r=1}^{10} r^2 \cdot {}^{10}C_r = 10 \left( (10-1)2^{10-2} + 2^{10-1} \right)$ $= 10 \left( 9 \cdot 2^8 + 2^9 \right)$ $= 10 \left( 9 \cdot 2^8 + 2 \cdot 2^8 \right) = 10 \left( (9+2) \cdot 2^8 \right)$ $= 10 \left( 11 \cdot 2^8 \right) = 110 \cdot 2^8$ We can rewrite $110 \cdot 2^8 = 55 \cdot 2 \cdot 2^8 = 55 \cdot 2^9$.

Now substitute these results back into the LHS expression: $LHS = 2 \left( 110 \cdot 2^8 \right) - \left( 10 \cdot 2^9 \right)$ $LHS = 220 \cdot 2^8 - 10 \cdot 2^9$ To simplify, convert $220 \cdot 2^8$ to $110 \cdot 2 \cdot 2^8 = 110 \cdot 2^9$: $LHS = 110 \cdot 2^9 - 10 \cdot 2^9$ $LHS = (110 - 10) \cdot 2^9 = 100 \cdot 2^9$ We can further simplify $100 \cdot 2^9 = 50 \cdot 2 \cdot 2^9 = 50 \cdot 2^{10} = 25 \cdot 2 \cdot 2^{10} = 25 \cdot 2^{11}$. So, the Left Hand Side simplifies to: $LHS = 25 \cdot 2^{11}$

3. Evaluating the Right Hand Side (RHS) Expression Term

The RHS contains a term in parentheses: ${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + \dots \text{ upto 10 terms}$ With $C_r = {}^{10}C_r$, this sum is of the form $\sum \frac{{}^{10}C_r}{r+1}$. A note on "upto 10 terms": If $C_0$ is the first term, then the 10th term would be $\frac{{}^{10}C_9}{10}$. This implies the sum is $\sum_{r=0}^{9} \frac{{}^{10}C_r}{r+1}$. However, the solution proceeds by using a formula for $\sum_{r=0}^{n} \frac{{}^{n}C_r}{r+1}$, effectively summing up to $r=10$ (11 terms). We will follow the solution's implicit interpretation that the sum extends to $\frac{{}^{10}C_{10}}{11}$. This is a common point of confusion in competitive exams where "upto N terms" might sometimes imply the natural range of the index in a known formula.

Derivation and Calculation of $\sum_{r=0}^{n} \frac{{}^n C_r}{r+1}$ Consider the binomial expansion of $(1+x)^n$: $(1+x)^n = \sum_{r=0}^n {}^n C_r x^r$ Integrate both sides with respect to $x$ from $0$ to $1$: $\int_0^1 (1+x)^n dx = \int_0^1 \left( \sum_{r=0}^n {}^n C_r x^r \right) dx$ Evaluate the integral on the LHS: $\left[ \frac{(1+x)^{n+1}}{n+1} \right]_0^1 = \frac{(1+1)^{n+1}}{n+1} - \frac{(1+0)^{n+1}}{n+1} = \frac{2^{n+1}-1}{n+1}$ Evaluate the integral on the RHS: $= \sum_{r=0}^n {}^n C_r \left[ \frac{x^{r+1}}{r+1} \right]_0^1 = \sum_{r=0}^n {}^n C_r \left( \frac{1^{r+1}}{r+1} - \frac{0^{r+1}}{r+1} \right)$ $= \sum_{r=0}^n \frac{{}^n C_r}{r+1}$ Equating both sides, we get the identity: $\sum_{r=0}^n \frac{{}^n C_r}{r+1} = \frac{2^{n+1}-1}{n+1}$ For $n=10$ (assuming the sum goes from $r=0$ to $r=10$): $\sum_{r=0}^{10} \frac{{}^{10}C_r}{r+1} = \frac{2^{10+1}-1}{10+1} = \frac{2^{11}-1}{11}$ So, the term in the parenthesis on the RHS simplifies to: $\left( {C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + \dots \text{ upto 10 terms} \right) = \frac{2^{11}-1}{11}$

Now, substitute this back into the overall RHS expression: $RHS = {{\alpha \cdot {2^{11}}} \over {{2^\beta } - 1}} \left( \frac{2^{11}-1}{11} \right)$

4. Equating LHS and RHS to Find $\alpha$ and $\beta$

We have calculated $LHS = 25 \cdot 2^{11}$. Equating LHS and RHS: $25 \cdot 2^{11} = {{\alpha \cdot {2^{11}}} \over {{2^\beta } - 1}} \left( \frac{2^{11}-1}{11} \right)$ We can divide both sides by $2^{11}$: $25 = \frac{\alpha}{2^\beta - 1} \left( \frac{2^{11}-1}{11} \right)$ To match the form for direct comparison, rearrange the terms slightly: $25 \cdot 11 = \alpha \left( \frac{2^{11}-1}{2^\beta - 1} \right)$ Comparing this with the form shown in the question and the original solution's comparison step, we can infer the grouping: $25 \cdot 2^{11} = 2^{11} \left( \frac{\alpha}{11} \right) \left( \frac{2^{11}-1}{2^\beta - 1} \right)$ For this equality to hold, we compare the multiplicative factors: First, for the terms containing $\alpha$: $\frac{\alpha}{11} = 25 \implies \alpha = 25 \cdot 11 \implies \alpha = 275$ Next, for the remaining terms: $\frac{2^{11}-1}{2^\beta - 1} = 1$ This implies: $2^{11}-1 = 2^\beta - 1$ Adding 1 to both sides: $2^{11} = 2^\beta$ Therefore, by comparing the exponents: $\beta = 11$

Finally, we need to find the value of $\alpha + \beta$: $\alpha + \beta = 275 + 11 = 286$

5. Tips and Common Mistakes

• Careful Interpretation of Summations: Pay close attention to the range of summation, especially when "upto N terms" is mentioned. In binomial series, $C_0$ is often the first term, so "upto 10 terms" might mean $r=0$ to $r=9$. However, some problems implicitly align with standard identity ranges (e.g., $r=0$ to $n$ for $n+1$ terms) if that leads to a simpler solution. Always verify your interpretation.
• Derive Summation Identities: For complex sums involving $rC_r$, $r^2C_r$, or $C_r/(r+1)$, it's safer and more robust to quickly re-derive them using differentiation or integration of the binomial expansion rather than memorizing all forms.
• Algebraic Simplification: Be meticulous with algebraic manipulation and power simplifications (e.g., $2 \cdot 2^8 = 2^9$). A small error can propagate.
• Recognize Patterns: Identifying the general term in a series (like $(2r-1)rC_r$ for the LHS) is the first crucial step.

6. Key Takeaway

This problem demonstrates the power of calculus-based methods (differentiation and integration) to evaluate complex sums involving binomial coefficients. By relating the binomial expansion to its derivatives and integrals, we can transform sums into simpler expressions. This approach, combined with careful algebraic manipulation, is essential for solving advanced problems in binomial theorem.