2. Find Missing terms in interpolation table

x	f(x)
0	0
1	?
2	8
3	15
4	?
5	35

Solution:
The value of table for `x` and `y`

x	0	1	2	3	4	5
y	0	?	8	15	?	35

Here we are given 4 values of y. We assume that y is polynomial of degree 3.
Hence `Delta^4 y_k=0`

`(E-1)^4 y_k=0`

`(E^4-4E^3+6E^2-4E+1) y_k=0`

taking `k=0`

`y_4-4y_3+6y_2-4y_1+y_0=0`

`:.b-4(15)+6(8)-4(a)=0`

`:.b-4a-12=0`

`:.b-4a=12`

taking `k=1`

`y_5-4y_4+6y_3-4y_2+y_1=0`

`:.35-4(b)+6(15)-4(8)=0`

`:.-4b+a+93=0`

`:.-4b+a=-93`

Now solving this 2 equations using substitution method
`b-4a=12`

and `-4b+a=-93`

`4b-a=93`

Suppose,
`-4a+b=12 ->(1)`

and `-a+4b=93 ->(2)`

Taking equation `(1)`, we have

`-4a+b=12`

`=>b=4a+12 ->(3)`

Putting `b=4a+12` in equation `(2)`, we get

`-a+4b=93`

`-a+4(4a+12)=93`

`=>-a+16a+48=93`

`=>15a+48=93`

`=>15a=93-48`

`=>15a=45`

`=>a=3 ->(4)`

Now, Putting `a=3` in equation `(3)`, we get

`b=4a+12`

`=>b=4(3)+12`

`=>b=12+12`

`=>b=24`

`:.b=24" and "a=3`

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