Core 2 AQA Maths AS: Key formulae and proofs

Power Rule

Prove that logax^n= nlogax

First of all note that loga of a is ALWAYS 1.

p=logax   

x= a^p

x^n= (a^p)^n= a^pn                           Put the power on both sides.

logax^n= loga(a^pn)                            loga(a)= 1 so pn remains

logax^n= pn                                      p=logax

logax^n= nlogax                                                        

Addition rule

Prove that logaxy= logax + logay

Again note that loga of a is ALWAYS 1.

(p= logax, q=logay) therefore x= a^p and y= a^q

xy= a^p multiplied by a^q= a^p+q                     Rules of indices

logaxy= loga(a^p+q)                               loga(a)=1, so p+q remains

logaxy= p+q

logaxy= logax +logay                             p=logax, q=logay

Subtraction Rule

Prove that logax/y= logax-logay

loga(a)=1, p=logax, q=logay       x= a^p, y= a^q

x/y= a^p (divided by) a^q= a^p-q              Rules of indices

logax/y= loga(a^p-q)                                loga(a)=1

logax/y= p-q

logax/y= logax-logay                               p=logax, q=logay

Sum of Arithmetic Progression

Prove that Sn= n/2[ 2a + (n-1)d ]                    

L= nth term= a + (n-1)d

Sn= a + a + d + a + 2d +....................+ L - d + L

Sn (backwards)= L + L - d + L - 2d ............... + d + a + a

Adding them together you will notice that for every +ve 'd' there is a -ve 'd'. Therefore all 'd's are cancelled out.

2Sn= n( a + L ),  Sn = n/2( a + L ) then write L in as the nth to get the formula at the top. n/2[ a + a + (n-1)d ]

Sum of Geometric Progression

Prove that Sn= a(1 - r^n) / (1 - r)

Sn= a + ar + ar^2 + ar^3 + ................................... ar^n-1 + ar^n

rSn= ar + ar^2 + ar^3 + ar^4....................................ar^n-1 + ar^n

Sn - rSn= a + ar^n                            Then factorise both sides

Sn(1 - r)= a(1 - r^n)

Sn= a(1 - r^n) / (1 - r)