(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



  • $\begingroup$ 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

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, $$5^k-1=(5-1)(1+5^1+5^2+5^3+......+5^{k-1}) = 4*(1+5^1+5^2+5^3+......+5^{k-1})$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.