Given
x_o = 6, x₁ = 7, x₂ = 10, x₃ = 12
y_o = 13, y₁ = 14, y₂ = 15, y₃ = 17
Using Lagrange's formula,
y(11) = \(13\frac { (4)(1)(-1) }{ (-1)(-4)(-6) } +14\frac { (5)(1)(-1) }{ (1)(-3)(-5) } +5\frac { (5)(4)(-1) }{ (4)(3)(-2) } +17\frac { (5)(4)(1) }{ (6)(5)(2) } \)
= 2.1666 - 4.6666 + 12.5 + 5.6666
= 15.6666
∴ y(x) = 15.6666

Let us calculate the number of men whose wages is less than Rs. 35 by using Newton's forward interpolation formula
x_o + nh = x ⇒ 30 + n (10) = 35
⇒ 10n = 35 - 30 = 5
⇒ n = \(\frac{5}{10}\) = 0.5
The difference table is

\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
\(y(35)=9+\frac { 0.5 }{ 1! } (30)+\frac { (0.5)(0.5-1) }{ 2 } (5)+\frac { (0.5)(0.5-1)(0.5-2) }{ 6 } (2)\)
= 9 + 15 - 0.6 + 0.1 = 24 (approximately)
∴ Number of men getting wages between Rs. 30 and Rs. 35 is y(35) - y(30) = 24 - 9 = 15.

From the graph, it is clear that when x = 27, the value of y is 24.8

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

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

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

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.

Since only four values are given,
(E -1)4 u₀ = 0
⇒ (E⁴ - 4E³ + 6E² - 4E + 1) u₀
= u₄ - 4 u₃ + 6 u₂ - 4 u₁ + u₀
⇒ 385 - 4(u₃) + 6(520) - 4 (556) + 560 =0
⇒ 385 - 4u₃ + 3120 - 2224 + 560 = 0
⇒ 1841 - 4u₃ = 0
⇒ 1841 = 4u₃
⇒ u₃ = \(\frac{1841}{4}\) = 460.25
∴ u₃ = 460.25

Given

 Find f(2.8)
Since the required value 2.8 is at the end of the table, apply Newton's backward interpolation formula.
x_n + nh = 2.8 ⇒ 3 + n (1) = 2.8
⇒  n = 2.8 - 3 = -0.2
The difference table is

Newton's backward interpolation formula is
y(x = x_n + nh) = \(\frac { n }{ 1! } { \triangledown y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }\)
⇒ y(2.8) = 34 + (-0.2) (23) + \(\frac { (-0.2)(-0.2+1) }{ 2 } (14)+\frac { (0.2)(-0.2+1)(-0.2+2) }{ 6 } \)(16)
⇒ y(2.8) = 34 - 4.6 + (-0.2) (0.8) (7) + (-0.2) (0.8) (1.8)
⇒ y(2.8) = 34 - 4.6 - 1.12- 0.288
⇒ y(2.8) = 27.992

The forward interpolation formula is
\({ y }_{ ({ x=x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
Here x₀ + nh = x ⇒ x₀ = 0, h = 1
∴ 0 + n = x ⇒ n = x.
The difference table is

\({ y }_{ (n=x) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
\(y=1+\frac { n }{ 1! } (1)+\frac { n(n-1) }{ 2! } (-2)+\frac { n(n-1)(n-2) }{ 6 } (12)\)
⇒ y = 1 + x + (x² - x)(-1) + x(x² - 3x + 2) (2)
⇒ y = 1 + x - x² + x + 2x³ - 6x² + 4x
⇒ y = 1+ 6x - 7x² + 2x³
Hence, the cubic polynomial is 2x³ - 7x²+ 6x + 1.

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)
To find y at x = 1946
\(\therefore\) x₀ + nh = 1946, x₀ = 1941, h = 10
1941 + n(10) = 1946 \(\Rightarrow\) n = 0.5

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\(​​{ y }_{ \left( x=1946 \right) }=20+\frac { 0.5 }{ 1! } (4)+\frac { 0.5(0.5-1) }{ 2! } (1)+\frac { 0.5(0.5-1)(0.5-2) }{ 3! } (1)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3) }{ 4! } (0)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3)(0.5-4) }{ 5! } (-9)\)
= 20+2-0.125+0.0625-0.24609
= 21.69 lakhs

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

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 Δ (f(x)) = f(x + h) -f(x)
Since h = 1, ∆ (f)) = f(x + 1) - f(x)
[∵ (x + 1)³ = x³ + 3x² + 3x + 1]
⇒ Δ (x³) = (x + 1)³ - x³

⇒ Δ (x³) = 3x² + 3x + 1