Question:

Step 1: Express the given sum using summation notation and apply the identity $r {}^{n}C_r = n {}^{n-1}C_{r-1}$. The given sum is $S = ({}^{30}C_1)^2+2({}^{30}C_2)^2+3({}^{30}C_3)^2+\cdots+30({}^{30}C_{30})^2$. We can write this as: $$S = \sum_{r=1}^{30} r ({}^{30}C_r)^2$$ We use the identity $r {}^{n}C_r = n {}^{n-1}C_{r-1}$. For $n=30$, this means $r {}^{30}C_r = 30 {}^{29}C_{r-1}$. Substituting this into the sum, we get: $$S = \sum_{r=1}^{30} \left(30 {}^{29}C_{r-1}\right) {}^{30}C_r$$ $$S = 30 \sum_{r=1}^{30} {}^{29}C_{r-1} {}^{30}C_r$$ Step 2: Apply the symmetry property of binomial coefficients to one of the terms. The symmetry property states that ${}^{n}C_r = {}^{n}C_{n-r}$. Applying this to the term ${}^{30}C_r$, we can rewrite it as ${}^{30}C_{30-r}$. Substituting this into the sum: $$S = 30 \sum_{r=1}^{30} {}^{29}C_{r-1} {}^{30}C_{30-r}$$ Step 3: Change the index of summation to align with Vandermonde's Identity. Let $k = r-1$. When $r=1$, $k=0$. When $r=30$, $k=29$. Thus, $r = k+1$. Substituting $k$ into the sum: $$S = 30 \sum_{k=0}^{29} {}^{29}C_k {}^{30}C_{30-(k+1)}$$ $$S = 30 \sum_{k=0}^{29} {}^{29}C_k {}^{30}C_{29-k}$$ Step 4: Apply Vandermonde's Identity. Vandermonde's Identity states that $\sum_{k=0}^{r} {}^{m}C_k {}^{n}C_{r-k} = {}^{m+n}C_r$. Comparing our sum $30 \sum_{k=0}^{29} {}^{29}C_k {}^{30}C_{29-k}$ with Vandermonde's Identity, we have $m=29$, $n=30$, and the sum of the lower indices is $k + (29-k) = 29$. The upper limit of summation for $k$ is $29$. Thus, the sum evaluates to: $$\sum_{k=0}^{29} {}^{29}C_k {}^{30}C_{29-k} = {}^{29+30}C_{29} = {}^{59}C_{29}$$ Therefore, the sum $S$ becomes: $$S = 30 \cdot {}^{59}C_{29}$$ Step 5: Express the result in terms of factorials and simplify. Using the definition ${}^{n}C_r = \dfrac{n!}{r!(n-r)!}$, we expand ${}^{59}C_{29}$: $$S = 30 \cdot \dfrac{59!}{29! (59-29)!}$$ $$S = 30 \cdot \dfrac{59!}{29! 30!}$$ We can express $30!$ as $30 \cdot 29!$: $$S = 30 \cdot \dfrac{59!}{29! (30 \cdot 29!)}$$ $$S = \dfrac{30 \cdot 59!}{30 \cdot (29!)^2}$$ $$S = \dfrac{59!}{(29!)^2}$$ Step 6: Equate the simplified sum to the given form and solve for $\alpha$. The problem states that the sum is equal to $\dfrac{\alpha\cdot60!}{(30!)^2}$. So we have: $$\dfrac{59!}{(29!)^2} = \dfrac{\alpha\cdot60!}{(30!)^2}$$ To solve for $\alpha$, we express $60!$ as $60 \cdot 59!$ and $(30!)^2$ as $(30 \cdot 29!)^2 = 30^2 \cdot (29!)^2$: $$\dfrac{59!}{(29!)^2} = \dfrac{\alpha \cdot (60 \cdot 59!)}{30^2 \cdot (29!)^2}$$ We can cancel $59!$ and $(29!)^2$ from both sides: $$1 = \dfrac{\alpha \cdot 60}{30^2}$$ $$1 = \dfrac{60\alpha}{900}$$ Multiply both sides by 900: $$900 = 60\alpha$$ Now, solve for $\alpha$: $$\alpha = \dfrac{900}{60}$$ $$\alpha = 15$$ Step 7: Conclude the final answer. The value of $\alpha$ is 15. This matches Option 3. The final answer is $\boxed{15}$.