# Answer on Other Math Question for jane doe

Question #5124

what is the next number in this sequence of numbers? 0,10,24,56,112,190

Expert's answer

x,,,,,,1,,,,,2,,,,3,,,,,,4,,,,,5,,,,,,6

.

f(x),,,,0,,,10,,,24...56,,,112,,,190,,,,,,,,base

.

d,,,,,,,,,,10,,,,14,,,32,,,56,,,78,,,,,,,,,,,,,1st degree

.

d,,,,,,,,,,,,,,,,,4,,,,,18,,,24,,,22,,,,,,,,,,,,,,2nd degree

.

d,,,,,,,,,,,,,,,,,,,,,14,,,,6,,,,,-2,,,,,,,,,,,,,,,,,3rd degree

.

d,,,,,,,,,,,,,,,,,,,,,,,,,,-8,,,,-8,,,,,,,,,,,,,,,,,,,4th degree

.

This defines f(x) = ax^4 +bx^3 + cx^2 + dx +e

.

(1) 0 = a(1)^4 +b(1)^3 +c(1)^2 +d(1)^1 +e

.

(2) 10 = a(2)^4 +b(2)^3 +c(2)^2 +d(2)^1 +e

.

(3) 24 = a(3)^4 +b(3)^3 +c(3)^2 +d(3)^1 +e

.

(4) 56 = a(4)^4 +b(4)^3 +c(4)^2 +d(4)^1 +e

.

(5) 112 =a(5)^4 +b(5)^3 +c(5)^2 +d(5)^1 +e

.

(6) 190 = a(6)^4 +b(6)^3 +c(6)^2 +d(6)^1 +e

.

now solving these simultaneously, using successive substitutions (and mult),

or matrix computation

.

Results in approx,,, f(x) = -.03 x^4 +1.49x^3 -7.32x^2 +21.93x-16.07

.

checking shows good agreement at low end, but about 70 % of actual at high end

.

making this correction,,,f(7) = 311 approx

.

Time does not permit any further computation, but perhaps this will give you a good start

.

f(x),,,,0,,,10,,,24...56,,,112,,,190,,,,,,,,base

.

d,,,,,,,,,,10,,,,14,,,32,,,56,,,78,,,,,,,,,,,,,1st degree

.

d,,,,,,,,,,,,,,,,,4,,,,,18,,,24,,,22,,,,,,,,,,,,,,2nd degree

.

d,,,,,,,,,,,,,,,,,,,,,14,,,,6,,,,,-2,,,,,,,,,,,,,,,,,3rd degree

.

d,,,,,,,,,,,,,,,,,,,,,,,,,,-8,,,,-8,,,,,,,,,,,,,,,,,,,4th degree

.

This defines f(x) = ax^4 +bx^3 + cx^2 + dx +e

.

(1) 0 = a(1)^4 +b(1)^3 +c(1)^2 +d(1)^1 +e

.

(2) 10 = a(2)^4 +b(2)^3 +c(2)^2 +d(2)^1 +e

.

(3) 24 = a(3)^4 +b(3)^3 +c(3)^2 +d(3)^1 +e

.

(4) 56 = a(4)^4 +b(4)^3 +c(4)^2 +d(4)^1 +e

.

(5) 112 =a(5)^4 +b(5)^3 +c(5)^2 +d(5)^1 +e

.

(6) 190 = a(6)^4 +b(6)^3 +c(6)^2 +d(6)^1 +e

.

now solving these simultaneously, using successive substitutions (and mult),

or matrix computation

.

Results in approx,,, f(x) = -.03 x^4 +1.49x^3 -7.32x^2 +21.93x-16.07

.

checking shows good agreement at low end, but about 70 % of actual at high end

.

making this correction,,,f(7) = 311 approx

.

Time does not permit any further computation, but perhaps this will give you a good start

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