Value of Weighted Sum of Squared Binomial Coefficients

To evaluate the given expression, we start with the expression for $X$: $X = \left(^{10}C_1\right)^2 + 2\left(^{10}C_2\right)^2 + 3\left(^{10}C_3\right)^2 + \dots + 10\left(^{10}C_{10}\right)^2 = \sum_{r=1}^{10} r \left(^{10}C_r\right)^2$

Step 1: Finding a closed-form expression for $X$

Consider the standard binomial expansion: (1+x)^{10} = \sum_{r=0}^{10} ^{10}C_r x^r

Differentiating both sides with respect to $x$: $10(1+x)^9 = \sum_{r=1}^{10} r \cdot ^{10}C_r x^{r-1} \quad \text{--- (1)}$

Next, consider the reverse expansion of $(x+1)^{10}$: (x+1)^{10} = \sum_{k=0}^{10} ^{10}C_{10-k} x^k = \sum_{k=0}^{10} ^{10}C_k x^{10-k} \quad \text{--- (2)}

Multiplying equations (1) and (2): 10(1+x)^{19} = \left( \sum_{r=1}^{10} r \cdot ^{10}C_r x^{r-1} \right) \left( \sum_{k=0}^{10} ^{10}C_k x^{10-k} \right)

We want to find the coefficient of $x^9$ on both sides:

• On the left-hand side: The coefficient of $x^9$ in $10(1+x)^{19}$ is: $\text{Coeff. of } x^9 = 10 \cdot ^{19}C_9$

• On the right-hand side: The exponent of $x$ in the product of terms is $(r-1) + (10-k) = r - k + 9$. For the term to have $x^9$, we require: $r - k + 9 = 9 \implies r = k$ Summing the product of coefficients where $r = k$ gives: $\sum_{r=1}^{10} r \left(^{10}C_r\right)^2 = X$

Equating both sides, we get: $X = 10 \cdot ^{19}C_9$

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Step 2: Calculating $\frac{1}{1430} X$

Substitute $X = 10 \cdot ^{19}C_9$ into the required expression: $\frac{1}{1430} X = \frac{10 \cdot ^{19}C_9}{1430} = \frac{^{19}C_9}{143}$

Now, expand $^{19}C_9$: $^{19}C_9 = \frac{19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

Note that $143 = 11 \times 13$. Dividing $^{19}C_9$ by $143$ cancels out $11$ and $13$ from the numerator: $\frac{^{19}C_9}{143} = \frac{19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 12}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

Now, simplify the factors in the numerator and denominator:

1. $\frac{18}{9 \times 2} = 1$
2. $\frac{14}{7} = 2$
3. $\frac{15}{5 \times 3} = 1$
4. $\frac{12}{6} = 2$

Substituting these back into the fraction: $\frac{^{19}C_9}{143} = \frac{19 \times 17 \times 16 \times 2 \times 2}{8 \times 4} = \frac{19 \times 17 \times 16 \times 4}{32}$

Since $\frac{16 \times 4}{32} = \frac{64}{32} = 2$: $\frac{1}{1430} X = 19 \times 17 \times 2 = 323 \times 2 = 646$

Final Answer:

646