how do i find a formula for the series 1, 7, 25, 103?

Answer 1

There are probably many formulas that would generate that sequence... probably even one less complex than this one:

f(x) = 8x³ - 42x² + 76x - 41

f(1) = 8(1)³ -42(1)² + 76(1) - 41
f(1) = 8(1) - 42(1) + 76(1) - 41
f(1) = 8 - 42 + 76 - 41
f(1) = 1

f(2) = 8(2)³ -42(2)² + 76(2) - 41
f(2) = 8(8) - 42(4) + 76(2) - 41
f(2) = 64 - 168 + 152 - 41
f(2) = 7

f(3) = 8(3)³ -42(3)² + 76(3) - 41
f(3) = 8(27) - 42(9) + 76(3) - 41
f(3) = 216 - 378 + 228 - 41
f(3) = 25

f(4) = 8(4)³ -42(4)² + 76(4) - 41
f(4) = 8(64) - 42(16) + 76(4) - 41
f(4) = 512 - 672 + 304 - 41
f(4) = 103

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If you want to stick with a recursive formula like the ones others have mentioned here, you'd have:

f(1) = 1,
f(2k) = 4[f(2k - 1)] + 3, for natural numbers k > 1, and
f(2k + 1) = 4[f(2k)] - 3, for natural numbers k > 1.

Answer 2

Louise is right assuming the 3rd finite difference is constant (i.e. 48).

Generally, you take differences between the terms and derive a formula using finite differences.

To verify Louise's formula, all you need to do is show that f(5)=289, i.e. 48+60+78+103=289

Answer 3

4x+3 or 4x-3..
dont know ..

Answer 4

it is 4x +3*(-1)^n where n=0,1,2,....
and x = 1,2,3....

Answer 5

(4x)+3, (4x)-3, (4x)+3...