Mr. Rothman is using the Binomial Theorem to find the binomial expansion of (a+b)7
Which of the following is a term in the expansion?
21a6b
21 A to the 6th power b
a4b3
A to the 4th power b cubed
21a2b5
21 A squared b to the 5th power
21a2b4
21 a squared b to the 4th power
The correct term in the expansion of (a+b)7 is 21a6b.
To find the binomial expansion of (a+b)^7 using the Binomial Theorem, we need to use the formula:
(a+b)^n = nC0 * a^n * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nC(n-1) * a^1 * b^(n-1) + nCn * a^0 * b^n
In this case, n = 7.
Let's go through each option to determine if it is a term in the expansion:
1) 21a^6b
To determine if this is a term, we need to check the values of 'a' and 'b'. In this case, 'a' is raised to the 6th power, and 'b' appears to the first power. This matches the form of one of the terms in the binomial expansion, namely nCr * a^(n-r) * b^r. Therefore, 21a^6b is a term in the expansion.
2) 21 A to the 6th power b
The question seems to describe the same term as the previous option but written with 'A' instead of 'a'. Since the binomial expansion is in terms of variables 'a' and 'b', this option is likely incorrectly written and doesn't match the form of the terms in the expansion. Therefore, 21 A to the 6th power b is not a term in the expansion.
3) a^4b^3
This is not the same form as the terms in the binomial expansion. In the binomial expansion, the powers of 'a' start from 'n' and decrease while the powers of 'b' start from 0 and increase. Here, 'a' is raised to the 4th power and 'b' to the 3rd power, which does not match the form of the terms. Therefore, a^4b^3 is not a term in the expansion.
4) 21a^2b^5
Similar to the first option, this term matches the form nCr * a^(n-r) * b^r, where 'a' is raised to the 2nd power and 'b' is raised to the 5th power. Therefore, 21a^2b^5 is a term in the expansion.
So, the terms in the expansion are 21a^6b and 21a^2b^5.