Find the term of the14th sequences 2,6,18

Mathematics

1 answer:

Yakvenalex [24]3 years ago

4 0

Answer:

The geometric sequence seems to be multiplying each term by a value of 3. Therefore, the 14th term would be 1458.

Step-by-step explanation:

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Which expression is a factor of 16x^2-81

krek1111 [17]

Answer:

Step-by-step explanation:

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2 years ago

In the expansion of (3a + 4b)8, which of the following are possible variable terms? Explain your reasoning. a2b3; a5b3; ab8; b8;

Fiesta28 [93]

We have to find the expansion of $(3a+4b)^{8}$

We will use binomial expansion to expand the given expression, which states that the expression $(a+b)^{n}$ is expanded as :

$(a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{2}a^{n-2}b^{2}+........^{n}C_{n}b^{n}$

Now expanding $(3a+4b)^{8}$ we get,

$(3a+4b)^{8}=^{8}C_{0}(3a)^{8}+^{8}C_{1}(3a)^{7}(4b)+^{8}C_{2}(3a)^{6}(4b)^{2}+^{8}C_{3}(3a)^{5}(4b)^{3}+^{8}C_{4}(3a)^{4}(4b)^{4}+^{8}C_{5}(3a)^{3}(4b)^{5}+^{8}C_{6}(3a)^{2}(4b)^{6}+^{8}C_{7}(3a)(4b)^{7}+^{8}C_{8}(4b)^{8}$

$(3a+4b)^{8}=^{8}C_{0}(3)^{8}a^{8}+^{8}C_{1}(3)^{7}(4)(a^{7}b)+^{8}C_{2}(3)^{6}(4)^{2}(a^{6}b^{2})+^{8}C_{3}(3)^{5}(4)^{3}(a^{5}b^{3})+^{8}C_{4}(3)^{4}(4)^{4}(a^{4}b^{4})+^{8}C_{5}(3)^{3}(4)^{5}(a^{3}b^{5})+^{8}C_{6}(3)^{2}(4)^{6}(a^{2}b^{6})+^{8}C_{7}(3)(4)^{7}(ab^{7})+^{8}C_{8}(4)^{8}(b^{8})$