C0n + ... + Cnn#9892

C0n + ... + Cnn (Editorial)

The formula of the binomial theorem is as follows:

(a + b)^{n} = i = 0 \sum n C_{n}^{i} a^{i} b^{n - i}

(a + b)^n = \sum_{i = 0}^n{C_n^ia^ib^{n-i}}

If we set $a = b = 1$ a = b = 1, then this relation takes the following form:

(1 + 1)^{n} = i = 0 \sum n C_{n}^{i} 1^{i} 1^{n - i}

(1 + 1)^n = \sum_{i = 0}^n{C_n^i1^i1^{n-i}}

2^{n} = i = 0 \sum n C_{n}^{i} = C_{n}^{0} + C_{n}^{1} + ... + C_{n}^{n}

2^n = \sum_{i = 0}^n{C_n^i} = C_n^0 + C_n^1 + ... + C_n^n

Thus, the indicated sum equals $2^{n}$ 2^n.

Example

If $n = 1$ n = 1, then $C_{1}^{0} + C_{1}^{1} = 1 + 1 = 2$ C_1^0 + C_1^1 = 1 + 1 = 2

If $n = 2$ n = 2, then $C_{2}^{0} + C_{2}^{1} + C_{2}^{2} = 1 + 2 + 1 = 4$ C_2^0 + C_2^1 + C_2^2 = 1 + 2 + 1 = 4

If $n = 3$ n = 3, then $C_{3}^{0} + C_{3}^{1} + C_{3}^{2} + C_{3}^{3} = 1 + 3 + 3 + 1 = 8$ C_3^0 + C_3^1 + C_3^2 + C_3^3 = 1 + 3 + 3 + 1 = 8

Algorithm realization

Read the input value of $n$ .

scanf("%lld", &n);

Compute and print the result, the value of $2^{n}$ 2^n.

res = 1LL << n;
printf("%lld\n", res);

Python realization

Read the input value of $n$ .

n = int(input())

Compute and print the result, the value of $2^{n}$ 2^n.

res = 1 << n
print(res)