The sum of the first nn terms of a sequence, SnS_n, is given by Sn=an3+bn2+cn+dS_n = an^3 + bn^2 + cn + d, where aa, bb, cc and dd are constants. It is given that S1=5S_1 = 5, S2=20S_2 = 20, S3=57S_3 = 57 and S4=128S_4 = 128. Find SnS_n.

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Worked solution

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1. Formulate a system of linear equations based on the given values of $S_n$.

S1=a(1)3+b(1)2+c(1)+d=5  ⟹  a+b+c+d=5S_1 = a(1)^3 + b(1)^2 + c(1) + d = 5 \implies a + b + c + d = 5
S2=a(2)3+b(2)2+c(2)+d=20  ⟹  8a+4b+2c+d=20S_2 = a(2)^3 + b(2)^2 + c(2) + d = 20 \implies 8a + 4b + 2c + d = 20
S3=a(3)3+b(3)2+c(3)+d=57  ⟹  27a+9b+3c+d=57S_3 = a(3)^3 + b(3)^2 + c(3) + d = 57 \implies 27a + 9b + 3c + d = 57
S4=a(4)3+b(4)2+c(4)+d=128  ⟹  64a+16b+4c+d=128S_4 = a(4)^3 + b(4)^2 + c(4) + d = 128 \implies 64a + 16b + 4c + d = 128

2. Solve the system of linear equations using a graphing calculator.

Using a graphing calculator to solve the system of 4 linear equations:
a=2a = 2
b=−1b = -1
c=4c = 4
d=0d = 0

3. State the final expression for $S_n$.

Sn=2n3−n2+4nS_n = 2n^3 - n^2 + 4n

Answer

Sn=2n3−n2+4nS_n = 2n^3 - n^2 + 4n

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