If $$\sum_{r=1}^{30} \frac{r^3 (^{30}C_r)^2}{^{30}C_{r-1}} = \alpha \times 2^{29}$$ , then $$\alpha$$ is equal to:

Step 1: Simplify the ratio of binomial coefficients

We begin by simplifying the ratio of the binomial coefficients $$\frac{^{30}C_r}{^{30}C_{r-1}}$$ . Using the property $$\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$$ , we substitute $$n=30$$ to get $$\frac{30-r+1}{r}$$ , which simplifies to $$\frac{31-r}{r}$$ .

Step 2: Substitute the simplified ratio into the sum

Now, we substitute the simplified ratio back into the original summation. The term $$\frac{(^{30}C_r)^2}{^{30}C_{r-1}}$$ can be rewritten as $$\frac{^{30}C_r}{^{30}C_{r-1}} \times ^{30}C_r$$ . This allows us to replace the ratio with the expression we just found.

Step 3: Simplify the term inside the summation

We can simplify the term inside the summation further. One $$r$$ from $$r^3$$ cancels with the $$r$$ in the denominator of $$\frac{31-r}{r}$$ , leaving us with $$r^2 (31-r) ^{30}C_r$$ .

Step 4: Expand and use the identity $$r \cdot ^nC_r = n \cdot ^{n-1}C_{r-1}$$

Expand the term inside the summation to get $$31r^2 {}^{30}C_r - r^3 {}^{30}C_r$$ . We will evaluate each part separately. We use the identity $$r \cdot ^nC_r = n \cdot ^{n-1}C_{r-1}$$ to simplify terms involving $$r \cdot ^{30}C_r$$ .

Step 5: Evaluate the sums using known identities

We use the identities $$\sum_{r=1}^{n} r^2 \cdot ^nC_r = n(n-1)2^{n-2} + n2^{n-1}$$ and $$\sum_{r=1}^{n} r^3 \cdot ^nC_r = n^2 2^{n-1} + n(n-1)2^{n-2}$$ . Substituting $$n=30$$ into these identities gives the values for the sums.

Step 6: Substitute the evaluated sums back into the expression

Now, we substitute the calculated values of the two summations back into the expression $$31 \sum_{r=1}^{30} r^2 {}^{30}C_r - \sum_{r=1}^{30} r^3 {}^{30}C_r$$ .

Step 7: Calculate the final value and find $$\alpha$$

Factor out $$30 \times 2^{28}$$ and perform the arithmetic. We then adjust the power of 2 to match the form $$\alpha \times 2^{29}$$ to find the value of $$\alpha$$ .

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