This is a question about the Binomial Theorem. According to it, the expansion of the binomial term (x+a)^n involves a series of terms. The generic term in the expansion is given by:

Tk+1 = (nCk)(a^k)(x^(n-k))

In our expansion, n equals 7, a equals (-1/x^2) and x equals (2x). Substituting this in the original formula:

Tk+1 = (7Ck)[(-1/x^2)^k](2x)^(7-k)

The value of k will start at 0 and go up to 7, so we will have 8 terms in total. Let's start with the first term. In the first term, k=0:

T1 = (7C0)[(-1/x^2)^0](2x)^(7-0) = 128x^7

For the second term, k=1 and so on:

T2 = (7C1)[(-1/x^2)^1](2x)^(7-1) = -448x^4

Doing this for the remaining terms, always increasing 1 in the value of k as we go along, the expansion results in:

2 months ago

## Answered By Leonardo F

This is a question about the Binomial Theorem. According to it, the expansion of the binomial term (x+a)^n involves a series of terms. The generic term in the expansion is given by:

Tk+1 = (nCk)(a^k)(x^(n-k))

In our expansion, n equals 7, a equals (-1/x^2) and x equals (2x). Substituting this in the original formula:

Tk+1 = (7Ck)[(-1/x^2)^k](2x)^(7-k)

The value of k will start at 0 and go up to 7, so we will have 8 terms in total. Let's start with the first term. In the first term, k=0:

T1 = (7C0)[(-1/x^2)^0](2x)^(7-0) = 128x^7

For the second term, k=1 and so on:

T2 = (7C1)[(-1/x^2)^1](2x)^(7-1) = -448x^4

Doing this for the remaining terms, always increasing 1 in the value of k as we go along, the expansion results in:

(2x - 1/x^2)^7 = 128x^7 - 448x^4 + 672x - 560/x^2 + 280/x^5 - 84/x^8 + 14/x^11 - 1/x^14