Question

Help...Find the first three terms in the expansion of

How can i solve the following question on binomial theorem...

If any one can help, thanks in advance :)

Answer

Note that, by the Binomial Theorem:

(x - 1)^7 = ∑ (-1)^k * C(7, k) * x^(7 - k) (from k=0 to 7)

(x + 2)^9 = ∑ 2^k * C(9, k) * x^(9 - k) (from k=0 to 9).

We see that the degree of the polynomial this product produces is x^16. There is only one way to produce the x^16 term. This is by letting k = 0 in the first product and k = 0 in the second product to produce the first term to be:

(-1)^0 * C(7, 0) * x^(7 - 0) * 2^0 * C(9, 0) * x^(9 - 0) = x^16.

To produce the x^15 term, there are two ways to make it.

i) k = 0 in product 1, k = 1 in product 2

==> (-1)^0 * C(7, 0) * x^(7 - 0) * 2^1 * C(9, 1) * x^(9 - 1)

= 18x^15.

ii) k = 1 in product 1, k = 0 in product 2

==> (-1)^1 * C(7, 1) * x^(7 - 1) * 2^0 * C(9, 0) * x^(9 - 0)

= -7x^15.

Adding these up yields the second term to be 18x^15 - 7x^15 = 11x^15.

For x^14, there are 3 ways to produce the term:

i) k = 2 in product 1, k = 0 in product 2

==> (-1)^2 * C(7, 2) * x^(7 - 2) * 2^0 * C(9, 0) * x^(9 - 0)

= 21x^14.

ii) k = 1 in product 1, k = 1 in product 2

==> (-1)^1 * C(7, 1) * x^(7 - 1) * 2^1 * C(9, 1) * x^(9 - 1)

= -126x^14.

iii) k = 0 in product 1, k = 2 in product 2

==> (-1)^0 * C(7, 0) * x^(7 - 0) * 2^2 * C(9, 2) * x^(9 - 2)

= 144x^14.

Adding these up yields 21x^14 - 126x^14 + 144x^14 = 39x^14.

Therefore, the first 3 terms are x^16 + 11x^15 + 39x^14.

I hope this helps!

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