Question

# Prove that the sum of the coefficients of the odd powers of $$x$$ is the expansion of $$(1 + x + x^{2} + x^{3} + x^{4})^{n - 1}$$, when $$n$$ is a prime number other than $$5$$, is divisible by $$n$$.

Solution

## $${ (1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }) }^{ n-1 }=1+{ c }_{ 1 }x+{ c }_{ 2 }{ x }^{ 2 }+{ c }_{ 3 }{ x }^{ 3 }+........(A)\\ { (1-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }) }^{ n-1 }=1-{ c }_{ 1 }x+{ c }_{ 2 }{ x }^{ 2 }-{ c }_{ 3 }{ x }^{ 3 }+..........(B)$$Putting $$x=1$$ and subtracting $$(i)$$ and $$(ii)$$$$\Rightarrow { 5 }^{ n-1 }-1=2({ c }_{ 1 }+{ c }_{ 3 }+{ c }_{ 5 }.......)\\ \Rightarrow { c }_{ 1 }+{ c }_{ 3 }+{ c }_{ 5 }......=\dfrac { { 5 }^{ n-1 }-1 }{ 2 }$$Using fermats theorem if $$N$$ is prime to $$p$$ then $${ N }^{ p-1 }-1$$ is divisible by $$p$$$$\Rightarrow { 5 }^{ n-1 }-1$$ is divisible by $$n$$ as $$n$$ is prime other then $$5$$Hence proved.Chemistry

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