Transcription of MATHEMATICAL FORMULAE Algebra
1 MATHEMATICAL FORMULAEA lgebra1. (a+b)2=a2+2ab+b2;a2+b2=(a+b)2 2ab2. (a b)2=a2 2ab+b2;a2+b2=(a b)2+2ab3. (a+b+c)2=a2+b2+c2+2(ab+bc+ca)4. (a+b)3=a3+b3+3ab(a+b);a3+b3=(a+b)3 3ab(a+b)5. (a b)3=a3 b3 3ab(a b);a3 b3=(a b)3+3ab(a b) b2=(a+b)(a b) b3=(a b)(a2+ab+b2) +b3=(a+b)(a2 ab+b2) bn=(a b)(an 1+an 2b+an 3b2+ +bn 1) :a:a::: :an=am+ nifm>n=1ifm=n=1an mifm<n;a2R;a6=013. (am)n=amn=(an)m14. (ab)n=an:bn15. ab n= ;a6= n=1an;an=1a Ifam=ananda6= 1;a6=0thenm=n20. Ifan=bnwheren6=0,thena= b21. Ifpx;pyare quadratic surds and ifa+px=py,thena= 0 andx=y22. Ifpx;pyare quadratic surds and ifa+px=b+pythena=bandx=y23.
2 Ifa;m;nare positive real numbers anda6=1,thenlogamn=logam+logan24. Ifa;m;nare positive real numbers,a6=1,thenloga mn =logam logan25. Ifaandmare positive real numbers,a6=1thenlogamn=nlogam26. Ifa;bandkare positive real numbers,b6=1;k6=1,thenlogba=logkalogkb27 . logba=1logabwherea;bare positive real numbers,a6=1;b6=128. ifa;m;nare positive real numbers,a6= 1 and if logam=logan,thenm=nTypeset byAMS-TEX229. ifa+ib=0 wherei=p 1, thena=b=030. ifa+ib=x+iy,wherei=p 1, thena=xandb=y31. The roots of the quadratic equationax2+bx+c=0;a6= 0 are b pb2 4ac2aThe solution set of the equation is( b+p 2a; b p 2a)where = discriminant =b2 4ac32.
3 The roots are real and distinct if > The roots are real and coincident if = The roots are non-real if < If and are the roots of the equationax2+bx+c=0;a6=0theni) + = ba= coe . ofxcoe . ofx2ii) =ca=constant termcoe . ofx236. The quadratic equation whose roots are and is (x )(x )= ( + )x+ = Sx+P=0whereS=Sum of the roots andP=Product of For an arithmetic progression ( ) whose rst term is (a) and the commondi erence is (d).i)nthterm=tn=a+(n 1)dii) The sum of the rst (n)terms=Sn=n2(a+l)=n2f2a+(n 1)dgwherel=last term=a+(n 1) For a geometric progression ( ) whose rst term is (a) and common ratiois ( ),i)nthterm=tn=a n ) The sum of the rst (n)terms:Sn=a(1 n)1 if <1=a( n 1) 1if >1=naif =1:39.
4 For any sequenceftng;Sn Sn 1=tnwhereSn=Sum of the rst (n) =1 =1+2+3+ +n=n2(n+1). =1 2=12+22+32+ +n2=n6(n+ 1)(2n+1). =1 3=13+23+33+43+ +n3=n24(n+1) !=(1):(2):(3):::::(n 1) !=n(n 1)! =n(n 1)(n 2)! =::::.45. 0! = (a+b)n=an+nan 1b+n(n 1)2!an 2b2+n(n 1)(n 2)3!an 3b3+ +bn;n>1.
