Give the first three terms of (2 + x)5 in ascending powers of x

To find the first three terms of the binomial expansion (2 + x)^5 in ascending powers of x, we use the binomial theorem which states that (a + b)^n = Σ (n choose k) * a^(n-k) * b^k from k=0 to n.

1. For the first term (k=0), we have (5 choose 0) * 2^5 * x^0 = 1 * 32 * 1 = 32.

2. For the second term (k=1), we have (5 choose 1) * 2^4 * x^1 = 5 * 16 * x = 80x.

3. For the third term (k=2), we have (5 choose 2) * 2^3 * x^2 = 10 * 8 * x^2 = 80x^2.

So, the first three terms of the expansion (2 + x)^5 in ascending powers of x are 32, 80x, and 80x^2.

To find the first three terms of the binomial expansion (2 + x)^5 in ascending powers of x, we use the binomial theorem which states that (a + b)^n = Σ (n choose k) * a^(n-k) * b^k from k=0 to n.

Here, a = 2, b = x, and n = 5.

1. For the first term (k=0), we have (5 choose 0) * 2^5 * x^0 = 1 * 32 * 1 = 32.

2. For the second term (k=1), we have (5 choose 1) * 2^4 * x^1 = 5 * 16 * x = 80x.

3. For the third term (k=2), we have (5 choose 2) * 2^3 * x^2 = 10 * 8 * x^2 = 80x^2.

So, the first three terms of the expansion (2 + x)^5 in ascending powers of x are 32, 80x, and 80x^2.