If $$\sum_{r=1}^{30}\f\frac{r^{2}({}^{30}C_{r})^{2}}{{}^{30}C_{r-1}}=\alpha \t\times 2^{29}$$, then $$\alpha$$ is equal to______.

We are given the summation: $$\sum_{r=1}^{30} \f\frac{r^{2} ({}^{30}C_{r})^{2}}{{}^{30}C_{r-1}} = \alpha \t\times 2^{29}$$ and need to find the value of $$\alpha$$.

First, recall the binomial coefficient definition: $${}^{n}C_{r} = \f\frac{n!}{r!(n-r)!}$$.

Consider the ratio $$\f\frac{{}^{30}C_{r}}{{}^{30}C_{r-1}}$$. Using the definitions:

$${}^{30}C_{r} = \f\frac{30!}{r! (30-r)!}$$ and $${}^{30}C_{r-1} = \f\frac{30!}{(r-1)! (31-r)!}$$.

So, $$\f\frac{{}^{30}C_{r}}{{}^{30}C_{r-1}} = \f\frac{\f\frac{30!}{r! (30-r)!}}{\f\frac{30!}{(r-1)! (31-r)!}} = \f\frac{(r-1)! (31-r)!}{r! (30-r)!}$$.

Simplify using $$r! = r \cdot (r-1)!$$ and $$(31-r)! = (31-r) \cdot (30-r)!$$:

$$\f\frac{(r-1)! \cdot (31-r) \cdot (30-r)!}{r \cdot (r-1)! \cdot (30-r)!} = \f\frac{31-r}{r}$$.

Thus, $$\f\frac{{}^{30}C_{r}}{{}^{30}C_{r-1}} = \f\frac{31-r}{r}$$.

Now, rewrite the given summation:

$$\f\frac{r^{2} ({}^{30}C_{r})^{2}}{{}^{30}C_{r-1}} = r^{2} \cdot ({}^{30}C_{r}) \cdot \f\frac{{}^{30}C_{r}}{{}^{30}C_{r-1}} = r^{2} \cdot ({}^{30}C_{r}) \cdot \f\frac{31-r}{r} = r(31-r) \cdot {}^{30}C_{r}$$.

So the sum becomes:

$$\sum_{r=1}^{30} r(31-r) {}^{30}C_{r}$$.

Expand the expression inside:

$$r(31-r) = 31r - r^2$$, so:

$$\sum_{r=1}^{30} (31r - r^2) {}^{30}C_{r} = 31 \sum_{r=1}^{30} r {}^{30}C_{r} - \sum_{r=1}^{30} r^2 {}^{30}C_{r}$$.

Use standard binomial identities. Recall that for a binomial expansion $$(1+x)^n = \sum_{r=0}^{n} {}^{n}C_{r} x^r$$:

1. The sum $$\sum_{r=0}^{n} r {}^{n}C_{r} = n \cdot 2^{n-1}$$.
2. The sum $$\sum_{r=0}^{n} r^2 {}^{n}C_{r} = n(n+1) 2^{n-2}$$.

Derive the second identity. Note that $$r^2 = r(r-1) + r$$, so:

$$\sum_{r=0}^{n} r^2 {}^{n}C_{r} = \sum_{r=0}^{n} r(r-1) {}^{n}C_{r} + \sum_{r=0}^{n} r {}^{n}C_{r}$$.

We know $$\sum_{r=0}^{n} r {}^{n}C_{r} = n \cdot 2^{n-1}$$.

For $$\sum_{r=0}^{n} r(r-1) {}^{n}C_{r}$$, note that for $$r=0,1$$, the term is zero, so start from $$r=2$$:

$$r(r-1) {}^{n}C_{r} = r(r-1) \f\frac{n!}{r!(n-r)!} = \f\frac{n!}{(r-2)!(n-r)!} = n(n-1) \f\frac{(n-2)!}{(r-2)!(n-r)!} = n(n-1) {}^{n-2}C_{r-2}$$.

Thus, $$\sum_{r=2}^{n} r(r-1) {}^{n}C_{r} = n(n-1) \sum_{r=2}^{n} {}^{n-2}C_{r-2} = n(n-1) \sum_{k=0}^{n-2} {}^{n-2}C_{k} = n(n-1) 2^{n-2}$$ (where $$k=r-2$$).

So, $$\sum_{r=0}^{n} r^2 {}^{n}C_{r} = n(n-1) 2^{n-2} + n \cdot 2^{n-1} = n(n-1) 2^{n-2} + 2n \cdot 2^{n-2} = n 2^{n-2} (n-1 + 2) = n(n+1) 2^{n-2}$$.

Now apply these to $$n=30$$. Note that the sums from $$r=0$$ include terms that are zero at $$r=0$$, so:

$$\sum_{r=1}^{30} r {}^{30}C_{r} = \sum_{r=0}^{30} r {}^{30}C_{r} = 30 \cdot 2^{29}$$,

$$\sum_{r=1}^{30} r^2 {}^{30}C_{r} = \sum_{r=0}^{30} r^2 {}^{30}C_{r} = 30 \cdot 31 \cdot 2^{28}$$.

Substitute back:

$$31 \sum_{r=1}^{30} r {}^{30}C_{r} - \sum_{r=1}^{30} r^2 {}^{30}C_{r} = 31 \cdot (30 \cdot 2^{29}) - (30 \cdot 31 \cdot 2^{28})$$.

Factor out common terms:

$$= 31 \cdot 30 \cdot 2^{29} - 31 \cdot 30 \cdot 2^{28} = 31 \cdot 30 \cdot 2^{28} (2 - 1) = 31 \cdot 30 \cdot 2^{28} \cdot 1 = 31 \cdot 30 \cdot 2^{28}$$.

The given equation is:

$$\sum_{r=1}^{30} \f\frac{r^{2} ({}^{30}C_{r})^{2}}{{}^{30}C_{r-1}} = \alpha \t\times 2^{29}$$.

So,

$$31 \cdot 30 \cdot 2^{28} = \alpha \t\times 2^{29}$$.

Solve for $$\alpha$$:

$$\alpha = \f\frac{31 \cdot 30 \cdot 2^{28}}{2^{29}} = \f\frac{31 \cdot 30}{2} = 31 \cdot 15 = 465$$.

Thus, $$\alpha = 465$$.