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# P6-pls. Prove proofs is easy watch

1. i am having mania with induction problems. i don't mind if that mania leads me to the correct answers but alas. i have attached a pdf file that conatins five of my difficulties. your help would be really appreciated. thanks.
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2. proof.pdf (25.2 KB, 96 views)
3. Here we go:

sin x = x - x³/3! + x^5/5! - ...

[x³/6 - x^5/120 + ...] = x - sin x

but for x>0, [x³/6 - x^5/120 + ...] > 0

therefore x - sin x > 0

x > sin x (1)

similarly

sin x = x - x³/6 + x^5/120 - x^7/5040 + ...

[x^5/120 - x^7/5040 + ...] = sin x - x + x³/6

for x > 0 , [x^5/120 - x^7/5040 + ...] > 0

therefore sin x - x + x³/6 > 0

sin x > x - x³/6 (2)

combine (1) and (2) gives

x > sin x > x - x³/6

as required
4. let the general (nth) term in a sequence be n³ + 6n² + 8n , denoted as Un
then Un+1 = (n+1)³ + 6(n+1)² + 8(n+1)
= n³ + 9n² + 23n + 15

subtracting Un from this:

Un+1 - Un = 3n² +15n +15
= 3[n² + 5n + 5]

therefore, the difference of two adjacent terms is always divisible by 3

therefore, if Ui is divisible by 3, then so is Ui+1 (1)

now U1 = 1 + 6 + 8 = 15 = 5x3

therefore U1 is divisible by 3, and by line (1), so is U2 , U3 etc

therefore n³ + 6n² + 8n is divisible by 3 for all positive integer values of n.
5. similar method as before, Un = 7^(2n) + [2^(3n -3)][3^(n-1)]

=7^(2n) + [{2^(3n)}{3^n}/24]

therefore 24Un = 24[7^(2n)] + [2^(3n)][3^n]

now, Un+1 = 49[7^(2n)] + [2^(3n)][3^n]

therefore Un+1 - 24Un = 25[7^(2n)]

which is divisible by 25.

Then follow the same argument as before

ie U1 = 50 = 2x25
6. (Original post by Faaip De Oiad)
similar method as before, Un = 7^(2n) + [2^(3n -3)][3^(n-1)]

=7^(2n) + [{2^(3n)}{3^n}/24]

therefore 24Un = 24[7^(2n)] + [2^(3n)][3^n]

now, Un+1 = 49[7^(2n)] + [2^(3n)][3^n]

therefore Un+1 - 24Un = 25[7^(2n)]

which is divisible by 25.

Then follow the same argument as before

ie U1 = 50 = 2x25
thank you thank you. i was wondering as to how to use them. and i can see why u are addicted. you are good at them
7. 98.
(i)
Let f(n) = 7^n - 6n - 1, and assume that f(k) is divisible by 36. Now:
f(k+1) = 7^(k+1) - 6k - 7 = 7.7^k - 6k - 7
And:
f(k+1) - f(k) = (7-1)7^k - 6 = 6(7^k - 1), thus f(k+1) is divisible by 6 iff f(k) is divisible by 6. (Check it for n=2 to conclude the induction.)

Now we want to prove that 7^k-1 is divisible by 6. Let g(n) = 7^n - 1, and assume as before that g(k) is divisible by 6. Now:
g(k+1) = 7^(k+1) - 1
And:
g(k+1) - g(k) = 6(7^k), thus g(k+1) is divisible by 6 iff g(k) is divisible by 6. (Again, check it for n=2 to complete the induction.)

Since f(k) = 6g(k), and g(k)=6m, then: f(k) = 6*6m = 36m. This concludes the proof.

(ii) Similar argument as above, except show that both are divisible by 4 and hence the expression is divisible by 4*4=16.

(iii)
We know that:
f(n) = 7^n - 6n - 1 is divisible by 36 (and hence by 4)
h(n) = 5^n - 4n - 1 is divisible by 16 (and hence by 4)

Now:
f(n) - h(n) = 7^n - 5^n - 2n

So: 7^n - 5^n - 2n is divisible by 4 since the LHS is divisible by 4. QED.

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