Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
(ie) Δ⁴y₀ = 0, ∴ (E−1)⁴y = 0
(E⁴ - 4E³ + 6E² - 4E +1)y₀ = 0
E⁴y₀ - 4E³y₀+ 6E²y₀ - 4Ey₀ +y₀  = 0
y₄ − 4y₃ + 6y₂ − 4y₁ + y₀ = 0
67.4 − 4y₃+ 6(54.1) − 4(4.2) + 45 = 0
240.2 = 4y₃ ∴ y₃ = 60.05

Given y₃ = 2, y₄ = −6, y₅ = 8, y₆ = 9 and y₇ = 17
Δ⁴y₃ = (E−1)⁴y₃
= (E⁴ − 4E³ + 6E² − 4E+1)y₃
= E⁴y₃ − 4E³y₃ + 6E²y₃− 4Ey₃ + y₃
= y₇ − 4y₆ + 6y₅ −4y₄+ y₃
= 17 – 4(9) + 6(8) –4(–6) + 2
= 17 – 36 + 48 + 24 + 2 = 55

Δ²U₀ = (E-1)⁴U₀
= (E⁴ − 4E³+ 6E²− 4E+1)U₀
= E⁴U₀ - 4E³U₀ + 6E²U₀− 4EU₀+ U₀
= U₄ − 4U₃ + 6U₂ − 4U₁ + U₀
= 29 − 4(28) + 6(21) − 4(11) + 1.
= 156 – 156 = 0

We know that f(4) − f(3) =  Δf(3)
f(4) − f(3) = Δf(3)
= Δ[f(2) + Δf(2)] ∵[f(3) - f(2) = Δf(2)]
= Δf(2) +Δ²f(2)
= Δf(2) + Δ² [f(1) + Δf(1)]
∴ f(4) = f(4) + Δf(2) +  Δ² f(1) +  Δ³f(1)

Let k be the sixth term of the series in the difference table.
First we find the forward differences

x	y	∆	∆2
1	8
		4
2	12		3
		7
3	19		3
		10
4	29		3
		13
5	42		k-55
		k-42
6	k

Given that the second differences are constant
∴ k – 55 = 3
k = 58
∴ the sixth term of the series is 58

y = f(x) = x³+2x+1 for x = 1,2,3,4,5

x	y	Δy	Δ2y	Δ3y	Δ4y
1	4
		9
2	13		12
		21		6
3	34		18		0
		39		6
4	73		24
		63
5	136

The Forward difference table is given below

x	y	Δy	Δ2y	Δ3y
0	0
		0.174
10	0.174		-0.001
		0.173		-0.001
20	0.347		-0.002
		0.171
30	0.518

(i) Δe^ax = e^a(x+h)−e^x
= e^aX.e^h-e^ax [∵ a^m+n = a^m.aⁿ]
= e^ax[e^h-1]
 (ii) Δ²e^x = Δ.[Δe^x]
= Δ[e^x+h - e^x]
= Δ[e^xe^h - e^x]
= Δe^x [e^h - 1]
= (e^h - 1)Δe^x 
= (e^h−1).(e^h−1).e^x
= (e^h−1)².e^x
(iii) Δ log x = log(x+h) − log x
= log \(\frac{x + h}{x}\)
= log \(\left( \frac { x }{ x } +\frac { h }{ x } \right) \)
= log \(\left( 1+\frac { h }{ x } \right) \)

Since five values are given, the polynomial which fits the data is of degree four.
Hence Δ⁵y_k = 0 (i.e) (E−1)⁵y_k = 0
i.e., (E⁵ - 5E⁴ + 10E³ - 10E² + 5E - 1)y_k = 0
E⁵y_k - 5E⁴y_k+ 10E³y_k- 10E²y_k+ 5Ey_k - y_k = 0 (1)
Put k = 0 in (1)
E⁵y₀ - 5E⁴y₀+ 10E³y₀- 10E²y₀+ 5Ey₀ - y₀ = 0
y₅ − 5y₄ + 10y₃ − 10y₂ +5 y₁ - y₀ = 0
y₅ − 5(350) +10y₃−10(260)+5(220)− 200 = 0
y₅ + 10y₃ = 3450 (2)

Put k = 1 in (1)
E⁵y₁ - 5E⁴y₁+ 10E³y₁- 10E²y₁+ 5Ey₁- y₀ = 0
y₆ − 5y₅ + 10y₄ −1 0y₃ − y₁ = 0
430  −5y₅ +10(350) −10y₃ + 5(260)− 220 = 0
5y₅+10y₃ = 5010 (3)
(3) – (2) ⇒ 4y₅ = 1560
y₅ = 390
From (1)  390 +10y₃ = 3450
10y₃ = 3450 – 390
y₃ ≅ 306

\(\Delta \)\(\left[ \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } \right] \)
By Partial fraction method
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\frac { A }{ x+3 } +\frac { B }{ x+2 } \)
\(A=\frac { 5x+12 }{ x+12 } [x=-3]=\frac { -15+12 }{ -1 } =\frac { -3 }{ -1 } =-3\)
\(B=\frac { 5x+12 }{ x+3 } \)[x = -2] \(=\frac { 2 }{ 1 } =2\)
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } = \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(\Delta \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\Delta \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(=\left[ \frac { 3 }{ x+1+3 } -\frac { 3 }{ x+3 } \right] +\left\{ \frac { 2 }{ x+1+2 } -\frac { 2 }{ x+2 } \right\} \)
\(=3\left[ \frac { 1 }{ x+4 } -\frac { 1 }{ x+3 } \right] +2\left[ \frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \right] \)
\(=\left[ \frac { -3 }{ (x+4)(x+3) } -\frac { 2 }{ (x+3)(2+3) } \right] \)
\(=\frac { -5x-14 }{ (x+2)(x+3)(x+4) } \)