Since 82 lies at the beginning of the table, we can use Newton's forward interpolation formula
\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
Also x₀ + nh = 82 ⇒ 75 + n(5) = 82 ⇒ 5n = 82 - 75 = 7
⇒ n = \(\frac75\) = 1.4
The difference table is

\(y=2459+\frac { 1.4 }{ 1! } (-441)+\frac { (1.4)(1.4-1) }{ 2! } (-397)+\frac { (1.4)(1.4-1)(1.4-2) }{ 3! } (457)\)
= 2459 - 617.4 - 111.6 - 25.592
y = 1704. 408 when x = 82.

Since x = 0.2 lies at the beginning of the table, use Newton's foward interpolation formula
\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
Here h = 1, x_o = 0, x = 0.2
⇒ x₀ + nh = 0.2 ⇒ 0 + n(1) = 0.2 ⇒ n = 0.2
The forward difference table is

x	y	Δy	∆2y	Δ3y	Δ4y
0	176
1	185	9
2	194	9	0
3	202	8	-1	-1
4	212	10	2	3	4

 ∴ y = 176 +\(\frac { 0.2 }{ 1! } (9)+\frac { (0.2)(0.2-1) }{ 2! } (0)+\frac { (0.2)(0.2-1)(0.2-2) }{ 3! } (-1)+\frac { (0.2)(0.2-1)(0.2-2)(0.3-3) }{ 4! } (4)\)

= 176 + 1.8 - 0.048 - 0.1344 = 177.6176
ஃ Hence when x = 0.2, y = 177.6176.

Given h = 1
LHS = (E⁻¹Δ) x³
= Δ(E^-1(x³))
= Δ(x - h)³  [∵ E^-1f(x) = f(x - nh)]
= Δ(1 - h)³ [∵ h = 1]
= (x - 1+ 1)³ - (x - 1)³ [∵ Δf(x) =f(x + h) - f(x)]
= x³ - (x - 1)³
= x³ - (x³ - 3x² + 3x - 1)
[∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= x³ - x³ + 3x² - 3x + 1
= 3x² - 3x + 1
= RHS
Hence proved

when x = 0, y = 0 + 0 + 0 - 1 ⇒ y = -1
when x = 1, y = 1³-1²+ 1-1 ⇒ y=0.
when x = 2, y = 2³ - 2² + 2 - 1 ⇒ y = 8 - 4 + 1 ⇒ y = 5
when x = 3, y = 3³ - 3²+ 3 - 1 ⇒ y = 27 - 9 + 2 ⇒ y = 20
when x = 4, y = 4³ - 4²+ 4 -1 ⇒ y= 64 -16 + 3 ⇒ y = 51
when x = 5, y = 5³ - 5²+ 5 - 1 ⇒ y = 125 - 25 + 4 ⇒ y = 104.
Hence, the forward difference table is
 x  y Δy Δ²y Δ³y Δ⁴y 0 -1             1       1 0   4         5   6   2 5   10   0     15   6   3 20   16   0     31   6   4 51   22         53       5 104        

Δ(log ax) = log (ax + h) -log (ax)
= log\(\left( \frac { ax+h }{ ax } \right) \)
=log \(\left( \frac { ax }{ ax } +\frac { h }{ ax } \right) \)
= log \(\left( 1+\frac { h }{ ax } \right) \)
∴ ∆ (log ax) = log \(\left( 1+\frac { h }{ ax } \right) \)

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

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

Given x_o = 0, x₁ = 3, x₂ = 5, x₃ = 6, x₄ = 8
y_o = 276, y₁ = 460, y₂ = 414, y₃ = 343, y₄ = 110
Lagrange's formula is
y = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 })({ x }_{ 0 }-{ x }_{ 4 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })({ x }_{ 1 }-{ x }_{ 3 })({ x }_{ 1 }-{ x }_{ 4 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 })({ x }_{ 2 }-{ x }_{ 4 }) } { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 })({ x }_{ 3 }-{ x }_{ 4 }) } { y }_{ 3 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 4 }-{ x }_{ 0 })({ x }_{ 4 }-{ x }_{ 2 })({ x }_{ 4 }-{ x }_{ 3 })({ x }_{ 4 }-{ x }_{ 4 }) } { y }_{ 4 }\)
⇒ 276 \(\frac { (1)(-1)(-2)(-4) }{ (-3)(-5)(-6)(-8) } +460\frac { (4)(-1)(-2)(-4) }{ (3)(-2)(-3)(-5) } +414\frac { (4)(1)(-1)(-4) }{ (5)(2)(-1)(-3) } +343\frac { (4)(1)(-1)(-2) }{ (6)(3)(1)(-2) } +110\frac { (4)(1)(-1)(-2) }{ (8)(5)(3)(2) } \)
⇒ y = -3.066 + 163.555 + 441.6 - 152.44 + 3.666
⇒ y = 453.311.

Using Newton's backward interpolation formula,
we can find y when x = 58.
∴ x + nh = x ⇒ 60 + n(5) = 58
⇒ 5n = 58 - 60 = -2
⇒ n = \(\frac{-2}{5}\)= -0.4
and y(58) = \({ y }_{ n }+\frac { n }{ n! } \nabla { y }_{ n }+\frac { n(n+1) }{ 2! } { \nabla }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n-2) }{ 3! } { \nabla }^{ 3 }{ (y }_{ n })+\) .....
The difference table is

\(y(58)=68.48+\frac { (-0.4) }{ 1! } (-6)+\frac { (0.4)(-0.4+1) }{ 2! } (2.84)+\frac { (-0.4)(-0.4+1)(-0.4+2) }{ 3! } (-1.16)+\frac { (-0.4)(-0.4+2)(-0.4+2)(-0.4+3) }{ 3! } (0.68)\)
=68.48 + (0.4)(6) + \(\frac { (-0.4)(0.6) }{ 2 } (2.84)+\frac { (-0.4)(0.6)(1.6) }{ 6 } (-1.16)+\frac { (-0.4)(0.6)(1.6)(2.6) }{ 24 } (0.68)\)
= 68.48 + 2.4 - 0.3408 + 0.07424 - 0.028288
= 70.5851052
⇒ y(58) = 70.59
∴ Hence, premium for a policy maluting at the age of 58 is 70.59.

\(\Delta \)\(\left[ \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } \right] \)
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] \(=\frac { 2 }{ 1 } =2\)
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } = \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(\Delta \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\Delta \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(=\left[ \frac { 3 }{ x+1+3 } -\frac { 3 }{ x+3 } \right] +\left\{ \frac { 2 }{ x+1+2 } -\frac { 2 }{ x+2 } \right\} \)
\(=3\left[ \frac { 1 }{ x+4 } -\frac { 1 }{ x+3 } \right] +2\left[ \frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \right] \)
\(=\left[ \frac { -3 }{ (x+4)(x+3) } -\frac { 2 }{ (x+3)(2+3) } \right] \)
\(=\frac { -5x-14 }{ (x+2)(x+3)(x+4) } \)

Since we apply the forward interpolation formula,last four values of f(x) are taken into consideration (Take the values from x = 3).
The forward interpolation formula is
\({ y }_{ (x={ x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{0 }+\frac { n(n-1)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
x₀ + nh = 3.2, x₀ = 3,y = 1
\(\therefore n=\frac { 1 }{ 5 } \)
The difference table is

x	y	\(\Delta y\)	\({ \Delta }^{ 2 }y\)	\({ \Delta }^{ 2 }y\)
3	16
		4
4	20		0
		4		10
5	24		10
6	38

y_(x=3.2) = \(16+\cfrac { 1 }{ 5 } (4)+\cfrac { \frac { 1 }{ 5 } \left( \frac { -4 }{ 5 } \right) }{ 2 } \left( 0 \right) +\cfrac { \frac { 1 }{ 5 } \left( \frac { -4 }{ 5 } \right) \left( \frac { -9 }{ 5 } \right) }{ 6 } \times 10\)
= 16+0.8+0+0.48
= 17.28

We know  f(x), ∇f(x + h), ∇f(x + 2h),... are the first order differences of f(x).
= f(x) - 2f(x - h) + f(x - 2h)
Consider ∇²f(x) = ∇[∇f(x))]
= ∇[f(x) - f(x - h)]
= ∇ f(X) - ∇ f(x - h)
= [f(x) - f(x - h)] - [f(x - h) - f(x - 2h)]
= f(x) - f(x - h) - f(x - h) - f(x - 2h)
∴ ∇² f(x), ∇² f(x + h), ∇² f(x + 2h) ... are the second order differences of f(x).

By data we have,
x₀ = 0, x₁ = 1, x₂ = 3
y₀ = 5, y₁ = 6, y₂ = 50 and x = 2
Using Lagrange's formula, we get
\(y=\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 0 }-{ x }_{ 1 }) } { y }_{ 2 }\)
= \(5\times \frac { (2-0)(2-3) }{ (0-1)(0-3) } +6\times \frac { (2-0)(2-3) }{ (1-0)(1-3) } +50\times \frac { (2-0)(2-1) }{ (3-0)(3-1) } \)

= \(\frac { 10 }{ 3 } -6+\frac { 50 }{ 3 } =\frac { 60 }{ 3 } -6=20-6=14\)
∴ f(2) = 14