Mathematical Induction (1 Viewer)

[mazza_728]

hey guys, i cant seem to prove this:
1x2^0+2x2^1+3x2^2+....+nx2^(n-1) = 1+(n-1)2^n

Ok the line im stuck on is LHS=1+(k-1)2^k+(k+1)2^k i need to prove that this
= 1+kx2^(k+1)

This question is from 1986 HSC Question 7 i) if anyone has this paper and has finished it... how did you do Question 7 ii)??
??
any help would be great
thanks xoxo

 

[Xayma]

Ok for the full induction:

Let n=1
LHS=1
RHS=1+0*2¹
=LHS

∴ true for n=1.

Assume true for n=k

1*2⁰+2*2¹+3*2²+......+k*2^k-1=1+(k-1)2^k

Prove true for n=k+1

1*2⁰+2*2¹+3*2²+......+k*2^k-1+(k+1)*2^k=1+(k-1)2^k+(k+1)*2^k

RHS=1+(k-1)*2^k+(k+1)*2^k
=1+2^k(k-1+k+1)
=1+2^k(2k)
=1+k*2*2^k
=1+k*2^k+1

∴ true for n=k+1 and since true for n=1, true for n=2,3,4.... and for all n≥1.

ii) a)
The height of the trapezium can be found to Lsinθ (as the upper angles are θ (alternate angles ll lines).

By forming triangles and a square the length of the top can be found to be L+2Lcosθ

∴ Area=1/2(2L+2Lcosθ.)Lsinθ
=L²(1+cosθ.)sinθ

b) dA/dθ=L²[(1+cosθ.).cosθ+sinθ.-sinθ]
=L²[cosθ+cos²θ-sin²θ]
=L²[cosθ+cos2θ]
For stat dA/dθ=0
cosθ+cos2θ=0
cosθ+2cos²θ-1=0
(2cosθ-1)(cosθ+1)=0
cosθ=-1 OR cosθ=1/2
θ=π OR cosθ=π/3

Now the first answer is beyond the range.

From here just test concavity to show that θ=π/3 gives a max.

 

[mazza_728]

Thanks Xayma ur a champ.