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Here’s how I solve this sort of problem.

First step: subtract the first term from each of the other values to produce a series with one fewer term: 19, 33, 38, 42, 45 and call these algebraically T₁ to T₅.

From these we are going to create 5 equations each containing 5 unknowns and we are basing them on the polynomial:

Tᵣ=a₅r⁵+a₄r⁴+a₃r³+a₂r²+a₁r, the coefficients being the unknowns to be found.

Second step: the primary equations:

① a₅+a₄+a₃+a₂+a₁=19 for r=1

② 32a₅+16a₄+8a₃+4a₂+2a₁=33 for r=2

③ 243a₅+81a₄+27a₃+9a₂+3a₁=38 for r=3

④ 1024a₅+256a₄+64a₃+16a₂+4a₁=42 for r=4

⑤ 3125a₅+625a₄+125a₃+25a₂+5a₁=45 for r=5

Third step: first reduction:

Subtract multiples of ① from each of the other equations using the rule Equation r minus r times Equation 1:

⑥ 30a₅+14a₄+6a₃+2a₂=-5

⑦ 240a₅+78a₄+24a₃+6a₂=-19

⑧ 1020a₅+252a₄+60a₃+12a₂=-34

➈ 3120a₅+620a₄+120a₃+20a₂=-50, which can be reduced to:

➈ 312a₅+62a₄+12a₃+2a₂=-5

Fourth step: second reduction:

Inspect the 4 equations above and manipulate them to eliminate another variable coefficient.

⑨-⑥: 282a₅+48a₄+6a₃=0

⑧-2⑦: 540a₅+96a₄+12a₃=4

Fifth step: third reduction:

We can see that doubling the first of these two equations gives us:

564a₅+96a₄+12a₃=0 and if we now subtract the second equation from this we get: 24a₅=-4, so a₅=-⅙.

Sixth step: substitution:

282a₅+48a₄+6a₃=0, -47+48a₄+6a₃=0, 48a₄+6a₃=47.

540a₅+96a₄+12a₃=4, -90+96a₄+12a₃=4, 96a₄+12a₃=94, which is the same as above and confirms the process.

Now continue the substitution of a₅=-⅙.

⑥ 30a₅+14a₄+6a₃+2a₂=-5⇒14a₄+6a₃+2a₂=0⇒7a₄+3a₃+a₂=0

⑦ 240a₅+78a₄+24a₃+6a₂=-19⇒78a₄+24a₃+6a₂=21⇒26a₄+8a₃+2a₂=7

⑧ 1020a₅+252a₄+60a₃+12a₂=-34⇒252a₄+60a₃+12a₂=136⇒63a₄+15a₃+3a₂=34

➈ 312a₅+62a₄+12a₃+2a₂=-5⇒62a₄+12a₃+2a₂=47

Seventh step: elimination of more unknowns:

⑦-⑥: 12a₄+2a₃=7

⑨-⑥: 48a₄+6a₃=47

⑨-⑥-3(⑦-⑥): 12a₄=26, a₄=13/6, a₃=½(7-26)=-19/2.

Eighth step: substitute all discovered coefficients: a₅, a₄, a₃.

a₂=-7(13/6)-3(-19/2)=40/3.

a₁=19-(-1/6+13/6-19/2+40/3)=79/6.

Tᵣ=-r⁵/6+13r⁴/6-19r³/2+40r²/3+79r/6.

Tᵣ+11=-r⁵/6+13r⁴/6-19r³/2+40r²/3+79r/6+11 produces the series:

11, 30, 44, 49, 53, 56, 30, -101, -481, -1342, -3024, -5995.

Please note that there will probably be other formulas generating a series with the given 6 terms.

by Top Rated User (1.1m points)

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