ohhh the fun..
FP1 - Mathematical induction to show....

(r=1 on bottom) Σ (n on top) r x 2(Power r) = 2(Power n - 1) + 2
(note: i will not be using r=1 throughout though you should)

Step 1 ----->> show true for n =1

LHS when n=1 is Σ (1 on top) r x 2(Power r) = 1 x 2(Power 1)
RHS when n=1 is 2(power 1+1)x(1-1) + 2... = 0 + 2... = 2

∴ RHS = LHS when n=1

step 2 and 3 can vary so be sure to practise all types of these questions..

Step 2 ------>> Assume true for n = k

Σ (K on top) r x 2(Power r) = 2(Power k+1) x (k-1) + 2

Step 3 ---->> using the assumption to show that it it is also true for n = k+1

LHS when n = k+1 .. ∑(k+1 on top) r x 2(Power r) which is.. ∑(k on top) + (k+1)th term

Now Using the assumption from the statement made the line above..

∑(k+1 on top) r x 2(Power r) ≡ ∑(k on top) r x 2(Power r) + (k+1) x 2(Power k+1)

≡ (( 2(Power k+1) x (k-1) + 2 )) + (( (k+1) x 2(Power k+1) ))
^ ^
^ ^
(Assumption made earlier with n=k) (( k+1)th term)

Now from this you may see it more visible writing down the powers for yourself but there is a common factor of 2(Power k+1)
so..

= 2(Power k+1)[ ( k-1) +( k+1) ] + 2

= 2(Power k+1) x [ 2k ] + 2
now taking the 2 out

= 2(Power k+2) x (k) + 2

now when we look at RHS when n = k+1

2(Power (k+1) + 1) x ( (k+1) - 1 ) + 2
2(Power k+2) x (k) + 2

Which is exactly the same as LHS when n = k+1
So this means also true for n = k so implies true for n =k+1,
therefore It will be true for all other positive integers by induction.

wow that was a mess so i decided to put it on comment, its not my fault i didnt realise the question wouldve been so long..

Awsome Drew!!!! Nice to see you getting involved. Have a look at your brothers work on the year 12 class page!

why ty =P