Suppose a, b are two real numbers and $$f(n)=$$α$$n+$$βn. LetΔ$$=31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)If$$ Δ$$=k($$α−$$1)2($$β−$$1)2($$α−β$$)2$$, then k is equal to

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Suppose a, b are two real numbers and $$f(n)=$$α$$n+$$βn. LetΔ$$=31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)If$$ Δ$$=k($$α−$$1)2($$β−$$1)2($$α−β$$)2$$, then k is equal to

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Δ$$=1+1+11+$$α$$+$$β$$1+$$α$$2+$$β$$21+$$α$$+$$β$$1+$$α$$2+$$β$$21+$$α$$3+$$β$$31+$$α$$2+$$β$$21+$$α$$3+$$β$$31+$$α$$4+$$β$$4=1111$$αβ$$1$$α$$2$$β$$21111$$αβ$$1$$α$$2$$β$$2=$$Δ$$12where$$ Δ$$1=1111$$αβ$$1$$α$$2$$β$$2Applying$$ $$C3$$→$$C3$$−$$C2$$ and $$C2$$→$$C2$$−$$C1$$ we getΔ$$1=1001$$α−$$1$$β−α$$1$$α$$2$$−$$1$$β$$2$$−α$$2=($$α−$$1)($$β−α$$)11$$α$$+1$$β$$+$$α$$=($$α−$$1)($$β−$$1)($$β−α$$)Thus$$ Δ$$=($$α−$$1)2($$β−$$1)2($$α−β$$)2$$∴ $$k=1$$

Suppose $a, b$ are two real numbers and $f (n) = α^{n} + β^{n}$ . Let
$Δ = |\begin{array}{ccc} 3 & 1 + f (1) & 1 + f (2) \\ 1 + f (1) & 1 + f (2) & 1 + f (3) \\ 1 + f (2) & 1 + f (3) & 1 + f (4) \end{array}|$
If $Δ = k (α - 1)^{2} (β - 1)^{2} (α - β)^{2},$ then $k$ is equal to

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Suppose a, b are two real numbers and f(n)=αn+βn. LetΔ=31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)If Δ=k(α−1)2(β−1)2(α−β)2, then k is equal to

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Suppose $a, b$ are two real numbers and $f (n) = α^{n} + β^{n}$ . Let
$Δ = |\begin{array}{ccc} 3 & 1 + f (1) & 1 + f (2) \\ 1 + f (1) & 1 + f (2) & 1 + f (3) \\ 1 + f (2) & 1 + f (3) & 1 + f (4) \end{array}|$
If $Δ = k (α - 1)^{2} (β - 1)^{2} (α - β)^{2},$ then $k$ is equal to