Find out the missing values in the following the data.

1.	Find out the missing values in the following the data.
Answer» Since 4 known values of y are given, then 4th leading difference may be zero. Δ⁴₀= (y – 1 )⁴ = y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0 (i) and the second equation can be obtained by increasing the suffixes of each term of ‘Y’ by one, keeping the coefficients same; we get: Δ⁴₁ = (y – 1)⁴ = y₅ – 4y₄ + 6y₃ – 4y₂ + y₁ = 0 …………. (ii) From equation (i) y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0 ie., 38 – 4(33) + 6y₂ – 4(20) + 13 = 0 So, by simplifying, 38 – 132 + 6y₂ – 80 + 13 = 0 ; 6y₂ – 161 We get y₂ =26.83 From (ii) y₅ – 4y₄ + 6y₃ – 4y₂ + y = 0 = 0 y₅ – 4(38) + 6(33) -4(26.83) + 20 = 0 y₅ – 152 + 198 – 107.32 + 20 = 0 ∴ y = 41.32.

Answer»

Since 4 known values of y are given, then 4th leading difference may be zero.

Δ⁴₀= (y – 1 )⁴ = y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0 (i)

and the second equation can be obtained by increasing the suffixes of each term of ‘Y’ by one, keeping the coefficients same; we get:

Δ⁴₁ = (y – 1)⁴ = y₅ – 4y₄ + 6y₃ – 4y₂ + y₁ = 0 …………. (ii)

From equation (i) y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0

ie., 38 – 4(33) + 6y₂ – 4(20) + 13 = 0

So, by simplifying, 38 – 132 + 6y₂ – 80 + 13 = 0 ;

6y₂ – 161

We get y₂ =26.83

From (ii) y₅ – 4y₄ + 6y₃ – 4y₂ + y = 0 = 0

y₅ – 4(38) + 6(33) -4(26.83) + 20 = 0

y₅ – 152 + 198 – 107.32 + 20 = 0

∴ y = 41.32.

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