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

Let x be the weight and y be the number of students.
Difference table of cumulative frequencies are given below

x	y	\(\Delta y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 3 }y\)	\(\Delta ^{ 4 }y\)
Below 40	250
		120
60	370		–20
		100
80	470
		70		10
100	540		–20
		50
120	590

Let us calculate the number of students whose weight is below 70. For this we use forward difference 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 = 70
\(\therefore\) x₀+nh = 70, x₀ = 40, h = 20
40+n(20) = 70 \(\Rightarrow\) n = 1.5
\({ y }_{ \left( x=70 \right) }=250+1.5\left( 120 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) }{ 2! } \left( -20 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) \left( -0.5 \right) }{ 3! } \left( -10 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) \left( -0.5 \right) \left( -1.5 \right) }{ 4! } \left( 20 \right) \)
= 250 + 180 - 7.5 + 0.625 + 0.46875
= 423.59
\(\cong \) 424.
Number of students whose weight is between
60 and 70 = y(70)−y(60) = 424−370 = 54

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

\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\Delta \left( \Delta \left( \frac { 1 }{ x } \right) \right) \)
Now \(\Delta \left[ \frac { 1 }{ x } \right] =\frac { 1 }{ 1+x } -\frac { 1 }{ x } \)
\({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) ={ \Delta }\left( \frac { 1 }{ 1+x } -\frac { 1 }{ x } \right) \)
\(=\Delta \left( \frac { 1 }{ 1+x } \right) -\Delta \left( \frac { 1 }{ x } \right) \)
Similarly \({ \Delta }^{ 2 }\left( \frac { 1 }{ x } \right) =\frac { 2 }{ x(x+1)(x+2) } \)

Since only five values of f(x) are given, the polynomial which fits the data is of degree 4.
Hence fifth differences are zeros.
∴ Δ⁵y₀ = 0
⇒ (E - 1)⁵y₀ = 0
⇒ (E⁵ - 5E⁴ + 10E³ - 10E² + 5E - 1) y₀ = 0
⇒ y₅ - 5y₄ + 10y₃ - 10y₂ + 5y₁ - y₀ = 0
⇒ 501 - 5(467) + 10y₃ - 10 (421) + 5 (391) - 363 =  0
⇒ 501 - 2335 + 10y₃ - 4210 + 1955 - 363 = 0
⇒ 10y₃ - 4452 = 0
⇒ 10y₃ = 4452
⇒ \({ y }_{ 3 }=\frac{4352}{10}\) = 445.2

Given f(x) = x²+ 3h
LHS = Δ f(x)
= f(x + h) - f(x)
= [(x + h)² + 3 (x + h)] - [x² + 3x]

= h² + 2xh + 3h
when h = 1,
LHS = 1² + 2x(1) + 3(1)
= 1+2x+3
= 2x+4= 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        

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