Question

The sum of the series $2 \times 1 \times {}^{20}C_4 - 3 \times 2 \times {}^{20}C_5 + 4 \times 3 \times {}^{20}C_6 - 5 \times 4 \times {}^{20}C_7 + ... + 18 \times 17 \times {}^{20}C_{20}$, is equal to

Answer 1

34

Answer 2

Let $S = 2 \times 1 \times {}^{20}C_4 - 3 \times 2 \times {}^{20}C_5 + 4 \times 3 \times {}^{20}C_6 - 5 \times 4 \times {}^{20}C_7 + \dots + 18 \times 17 \times {}^{20}C_{20}$.

We can write the general term of the series as $T_n = (-1)^{n-4} (n-2)(n-3) {}^{20}C_n$, where $n$ ranges from $4$ to $20$. So, $S = \sum_{n=4}^{20} (-1)^{n-4} (n-2)(n-3) {}^{20}C_n$.

Consider the expansion of $(1-x)^{20}$: $(1-x)^{20} = \sum_{k=0}^{20} (-1)^k {}^{20}C_k x^k$.

Differentiate twice with respect to $x$: First derivative: $\frac{d}{dx} (1-x)^{20} = -20(1-x)^{19} = \sum_{k=1}^{20} (-1)^k k {}^{20}C_k x^{k-1}$.

Second derivative: $\frac{d^2}{dx^2} (1-x)^{20} = -20(-19)(1-x)^{18} = 20 \times 19 (1-x)^{18}$. On the right side: $\sum_{k=2}^{20} (-1)^k k(k-1) {}^{20}C_k x^{k-2}$.

So, $20 \times 19 (1-x)^{18} = \sum_{k=2}^{20} (-1)^k k(k-1) {}^{20}C_k x^{k-2}$.

Let $j = k-2$. Then $k = j+2$. When $k=2$, $j=0$. When $k=20$, $j=18$. $20 \times 19 (1-x)^{18} = \sum_{j=0}^{18} (-1)^{j+2} (j+2)(j+1) {}^{20}C_{j+2} x^j$. Since $(-1)^{j+2} = (-1)^j$, we have: $20 \times 19 (1-x)^{18} = \sum_{j=0}^{18} (-1)^j (j+2)(j+1) {}^{20}C_{j+2} x^j$.

Let's use the identity $k(k-1) {}^{N}C_k = N(N-1) {}^{N-2}C_{k-2}$. Consider the sum $S = \sum_{n=4}^{20} (-1)^{n-4} (n-2)(n-3) {}^{20}C_n$. Let $j=n-4$. Then $n=j+4$. $S = \sum_{j=0}^{16} (-1)^j (j+2)(j+1) {}^{20}C_{j+4}$.

Let's use the general form: $\sum_{k=r}^n (-1)^{k-r} f(k) {}^{N}C_k$. Consider the identity $\sum_{k=r}^N (-1)^k {}^{N}C_k P(k) = 0$ if $P(k)$ is a polynomial of degree less than $N-r$. This is not directly applicable.

Let's recall the identity: $\sum_{k=0}^{n} (-1)^k {}^{n}C_k = 0$ for $n \ge 1$. $\sum_{k=0}^{n} (-1)^k k {}^{n}C_k = 0$ for $n \ge 2$. $\sum_{k=0}^{n} (-1)^k k(k-1) {}^{n}C_k = 0$ for $n \ge 3$. In general, $\sum_{k=0}^{n} (-1)^k k(k-1)\dots(k-m+1) {}^{n}C_k = 0$ for $n \ge m+1$.

Let's consider the sum $S = \sum_{n=4}^{20} (-1)^{n-4} (n-2)(n-3) {}^{20}C_n$. Let $k=n-4$. Then the sum becomes $S = \sum_{k=0}^{16} (-1)^k (k+2)(k+1) {}^{20}C_{k+4}$.

Let's use the property: $k(k-1) \dots (k-m+1) {}^{N}C_k = N(N-1) \dots (N-m+1) {}^{N-m}C_{k-m}$. We have $(n-2)(n-3) {}^{20}C_n$.

Now, let's consider the sum $S_{full} = \sum_{n=0}^{20} (-1)^n (n-2)(n-3) {}^{20}C_n$. $S_{full} = \sum_{n=0}^{20} (-1)^n [n(n-1) - 4n + 6] {}^{20}C_n$. $S_{full} = \sum_{n=0}^{20} (-1)^n n(n-1) {}^{20}C_n - 4 \sum_{n=0}^{20} (-1)^n n {}^{20}C_n + 6 \sum_{n=0}^{20} (-1)^n {}^{20}C_n$.

For $N=20$:

1. $\sum_{n=0}^{20} (-1)^n {}^{20}C_n = 0$ (since $20 \ge 1$).
2. $\sum_{n=0}^{20} (-1)^n n {}^{20}C_n = 0$ (since $20 \ge 2$).
3. $\sum_{n=0}^{20} (-1)^n n(n-1) {}^{20}C_n = 0$ (since $20 \ge 3$).

Therefore, $S_{full} = 0 - 4(0) + 6(0) = 0$.

Now, let's relate $S_{full}$ to the given series $S$. The given series is $S = \sum_{n=4}^{20} (-1)^{n-4} (n-2)(n-3) {}^{20}C_n$.

For $n=0,1,2,3$, the terms in $S_{full}$ are: For $n=0$: $(-1)^0 (0-2)(0-3) {}^{20}C_0 = 1 \times 6 \times 1 = 6$. For $n=1$: $(-1)^1 (1-2)(1-3) {^{20}C_1} = -1 \times (-1) \times (-2) \times 20 = -40$. For $n=2$: $(-1)^2 (2-2)(2-3) {^{20}C_2} = 1 \times 0 \times (-1) \times {^{20}C_2} = 0$. For $n=3$: $(-1)^3 (3-2)(3-3) {^{20}C_3} = -1 \times 1 \times 0 \times {^{20}C_3} = 0$.

So, $S_{full} = 6 - 40 + 0 + 0 + \sum_{n=4}^{20} (-1)^n (n-2)(n-3) {}^{20}C_n = 0$. $-34 + \sum_{n=4}^{20} (-1)^n (n-2)(n-3) {}^{20}C_n = 0$. So, $\sum_{n=4}^{20} (-1)^n (n-2)(n-3) {}^{20}C_n = 34$.

Our given series is $S = \sum_{n=4}^{20} (-1)^{n-4} (n-2)(n-3) {}^{20}C_n$. The signs in $S$ are $(-1)^{n-4}$. The signs in the sum we just calculated are $(-1)^n$. Since $(-1)^{n-4} = (-1)^n (-1)^{-4} = (-1)^n \times 1 = (-1)^n$. So, the signs are the same for $n \ge 4$.

Therefore, $S = \sum_{n=4}^{20} (-1)^n (n-2)(n-3) {}^{20}C_n = 34$.