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(b) (i) Use the identity $A^2-B^2=(A-B)(A+B)$ to factorise the expression $5^{2k}-1$.

Do I just put the k as 1 so that the equation is 5^2 and 1^2

Thanks

Steve

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  • $\begingroup$ No, there is no reason to think $k=1$. $\endgroup$ – Robert Israel Apr 11 '16 at 19:47
  • $\begingroup$ You can think of $5^{2k}$ as $(5^k)^2$ $\endgroup$ – Bernard Masse Apr 11 '16 at 19:50
  • $\begingroup$ Hi Thanks for the help so would the factor be simply (5^k-1)(5^k+1) $\endgroup$ – Steve Richrds Apr 11 '16 at 20:26
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For all real numbers A and B, $$A^2-B^2=(A-B)(A+B)$$ Substituting $$A=5^k$$ and $$B=1$$, $$(5^k)^2-1^2=(5^k-1)(5^k+1) \implies 5^{2k}-1= (5^k-1)(5^k+1)$$ Note that 5^k-1 can be further factorised if k is a natural number. In that case, $$5^k-1=(5-1)(1+5^1+5^2+5^3+......+5^{k-1}) = 4*(1+5^1+5^2+5^3+......+5^{k-1})$$

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  • $\begingroup$ @Steve If you are satisfied by the answer, you may consider accepting it. $\endgroup$ – Shubham Avasthi Apr 30 '16 at 7:58

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