Prove the expressions by induction method

[salubadsha]

hey guys

1st week of my university and i got the painful assignment. I've not touched math from last 4-5 months, so I'm having trouble with few questions. Please help me out, thanks in advance

1- Prove that if x not equal 0, then x + 2x^{2} + 3x^{3} + ... + nx^{n}(1+x) = [ x - (n - 1)x^{n+1} + nx^{n+2} ] / (1-x)^{2} for every poitive integer n.
2-Prove by induction that if x \geq 0 (1+x) ^n \geq 1+nx
3- Using the fact that d/dx(x) = 1 and the product rule, prove by induction that d/dx(x^{n}) = nx^{n-1} for every poitive integer n.
4- (a) Treating the equation 4x^2 +4xy +2y^2 − 4x − 2y +1 = 0 as a quadratic equation in x with coefficients in terms of y, solve for x.
b) Disprove the statement “For all real numbers x and y, 4x^2 +4xy +2y^2 − 4x − 2y +1 = 0”.

1st question: i tried n = 1 & then n = k but it didn't work out
2nd question: here's what i've

n = 0 (1 + x)^0 ≥ 1 + 0x. And that's true!
assume the expression is true when n = k
(1 + x)^k ≥ 1 + kx.
Prove for n = k + 1
(1 + x)^k + 1 ≥ 1 + (k + 1)x.
So (1 + x)^k + 1 = (1 + x)^k (1 + x)
From the 1st hypothesis
Therefore, (1 + x)^k (1 + x) ≥ (1 + kx) (1 + x)
after this I've no idea how to solve the rest of it

3rd question: no clue at all how to get started

4th question part a: since we need to put the who expression into ax^2 + bx + c so this is what i did

4x^2 + 4(y-1)x + 2y^2 - 2y + 1 = 0
therefore a = 4, b = 4(y-1) and c = 2y^2 - 2y + 1. I tried to use the quadratic formula but it doesn't give me a real root. Maybe i did something wrong and for part b i don't know how to solve it.

Please guys help me out as soon as possible, these question are due withing few days. All the help would be really apperciated , again tons of thanks in advance!

 

[HallsofIvy]

I don't understand what
x + 2x^{2} + 3x^{3} + ... + nx^{n}(1+x) = [ x - (n - 1)x^{n+1} + nx^{n+2} ] / (1-x)^{2}
is supposed to mean. In particular, what is the left side if n= 1? Do you really have that (1+x) term at the end?

2nd question: here's what i've

n = 0 (1 + x)^0 ≥ 1 + 0x. And that's true!
assume the expression is true when n = k
(1 + x)^k ≥ 1 + kx.
Prove for n = k + 1
(1 + x)^k + 1 ≥ 1 + (k + 1)x.
So (1 + x)^k + 1 = (1 + x)^k (1 + x)
From the 1st hypothesis
Therefore, (1 + x)^k (1 + x) ≥ (1 + kx) (1 + x)
after this I've no idea how to solve the rest of it

What is (1+ kx)(1+ x)? Multiply it. (And be careful about parentheses. (1+ x)^k+ 1 is not the same as
(1+ x)^(k+1)!)

3- Using the fact that d(x)/dx= 1 and the product rule, prove by induction that d(xⁿ)/dx= n x^n-1 for every poitive integer n.

It says "by induction" doesn't it? When n= 1, what is d(xⁿ)/dx= n x^n-1? If you know d(x^k)/dx= k x^k-1 for some k, what is d(x^k+1)dx? (Hint: the problem says use the product rule: x^k+1= x(x^k).

 

[salubadsha]

I don't understand what
x + 2x^{2} + 3x^{3} + ... + nx^{n}(1+x) = [ x - (n - 1)x^{n+1} + nx^{n+2} ] / (1-x)^{2}
is supposed to mean. In particular, what is the left side if n= 1? Do you really have that (1+x) term at the end?

nope sorry my bad, the correct statement is this:
x + 2x^{2} + 3x^{3} + ... + nx^{n}= [ x - (n - 1)x^{n+1} + nx^{n+2} ] / (1-x)^{2}

What is (1+ kx)(1+ x)? Multiply it. (And be careful about parentheses. (1+ x)^k+ 1 is not the same as
(1+ x)^(k+1)!)

I did mutliply it and this is what i got: 1 + x + kx + kx^2 but then i don't know how to proof that (1+x)^k(1+x) ≥ 1 + (k + 1)x

It says "by induction" doesn't it? When n= 1, what is d(xⁿ)/dx= n x^n-1? If you know d(x^k)/dx= k x^k-1 for some k, what is d(x^k+1)dx? (Hint: the problem says use the product rule: x^k+1= x(x^k).

if i continue from here, i'll get
d(x^k+1)/dx = (k + 1)(x<sup>k[/sup)

L.S. = d [(xk)(x)]/dx now if sub the d(xk)/dx value which is kxk-1 then
L.S. = kxk-1 . d(x)/dx Is this correct?
If the last statement is correct then could you please tell me how to solve the rest of it? Even if i use the definition of derivative i can't get the L.S. = R.S.

Any hints for question # 4?

Please someone help me with these, thanks!</sup>

 

[dedashton]

for question 4 the answer isn't a real root (should be x=(1-y(1+-i))/2
for part b you need to find a value of y such that y(1+-i) is real so y must be 0 and x is 1/2