Some help with maths (1 Viewer)

[tehsky]

Teacher gave our class like 20hrs of maths h/w so yeh :S
i know how to do most questions, just some of these i had problems with anyone wanna be so kind nuff to help out? ;O
The answers are in tha brackets i.e [answer]

1) The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and the 7th term is 46. Find the sum of the first 20 terms. [1010]

2) The 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the first 20 terms. [1290]

3) The sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. Find the 50th term of the series. [16]

4) Find the sum of all integers between 1 and 100 that are not multiples of 6. [4234]

5) Prove by mathematical induction: 4^n greater than or equal to 3n+7 for all integers n >1

6) Prove by mathematical induction: n(n+2) is divisible by 4 if n is any even positive integer

7) Evaluate: Definite integral where b = 4 and a = 2,
x^2
____dx [56/9]
3


8) Evaluate: Definite integral where b=4, a= 3:
x^2+x+3
_______ [0.0126]
3x^5


Cheers if you can help me out ;D It would be much appreciated

 

[tehsky]

Any help please? ;o

 

[watatank]

I hope you are following, I am skipping steps here....

1) The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and the 7th term is 46. Find the sum of the first 20 terms. [1010]

S(4) = 42 ---> 4/2 (2a + 3d) = 42 -----> 2a + 3d = 21 (equation 1)

T(3) + T(7) = 46 ---> (a + 2d) + (a + 6d) = 46 ---> 2a + 8d = 46 (equation 2)

Solve Equation 1 and 2 simultaneously, first by elimination...

2a + 3d = 21
2a + 8d = 46 goes to 5d=25 so d=5

Then 2a +3(5) = 21 ----> 2a + 15 =21 ---> 2a= 6 ---> a =3

S(20) = 20/2 [ 2.(3) + 19(5) ] = 1010 (by calc)

 

[watatank]

2) The 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the first 20 terms. [1290]

T(20) = 131 ---> a + 19d = 131 (equation 1)

T(6) + T(7) + T(8) + T(9) + T(10) = 235

(a + 5d) + (a + 6d) + (a + 7d) +(a + 8d) +(a + 9d) = 235

5a + 35d = 235 ---> a + 7d = 47 (Equation 2)

so like question one, solve simultaneously...by elmination

a + 19d = 131
a + 7d = 47

12d = 84 d = 7

a + 7(7)= 47

a = -2

S(20) = 20/2 [2(-2) + 19(7) ] = 1290

 

[watatank]

3) The sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. Find the 50th term of the series. [16]

249 - 233 = 16 need we say more?

...

Well since the sum of 50 terms is the sum of 49 terms plus the 50th term, the 50th term must be the difference of the sum of 50 terms and the sum of 49 terms....which is why the answer is the above..

 

[watatank]

4) Find the sum of all integers between 1 and 100 that are not multiples of 6. [4234]

First do the sum of all intergers

a = 1, d = 1, n = 100

S(100) = 100/2 [ 2(1) + 99(1)] = 5050

The sum of all intergers 1-100 = 5050

The sum of the multiples of 6 from 1-100

a = 6, l = 96, n = 16

S(16) = 16/2(6+96) = 816

The sum of the multiples of 6 from 1-100 is 816

Answer: 5010 - 816 = 4234

 

[watatank]

5) Prove by mathematical induction: 4^n greater than or equal to 3n+7 for all integers n >1

For n = 2

LHS = 4^2 = 16
RHS = 3(2) + 7 = 13 > LHS

True for n = 2

Assume for some k > 1

4^k greater than or equal to 3k+7.....or 4^k = 3k+7 + Q (Q is some postive number)

Then for n = k+1

LHS = 4^(k+1) = 4.(4^k) = 4.(3k+7 + Q)

which is greater or equal to (3k+7 + Q)

True for n = k+1

By mathematical induction, true for all integers > 1

 

[watatank]

7) Evaluate: Definite integral where b = 4 and a = 2,
x^2
____dx [56/9]
3

x^2
____dx = 1/3 x [ integral (x^2 dx), limit 2 to 4 ]
3

= 1/3 [ (x^3)/3 ], limit 2 to 4

= 1/3 [ 64/3-8/3]

= 1/3 [56/3]

= 56/9

 

[watatank]

6) Prove by mathematical induction: n(n+2) is divisible by 4 if n is any even positive integer

8) Evaluate: Definite integral where b=4, a= 3:
x^2+x+3
_______ [0.0126]
3x^5

Couldn't get these two. Too much thinking on my behalf is required for these ones and I dont want to think at this present time...I've never seen anything like question 8 either, perhaps 4 unit methods are needed... :s. Good luck!

 

[STx]

for q8 just divide the each of the terms in the numerator by the denominator then evaluate it normally

 

[tehsky]

for q8 just divide the each of the terms in the numerator by the denominator then evaluate it normally

you cant do that since there is no common term, the x is only there in x^2 and x , the 3 is x-less
and thanks alot for your help there mate
those two questions stumped me like hell :S worked on them for ages and zit. nada =\

 

[SoulSearcher]

(x²+x+3)/x⁵ = 1/x³ + 1/x⁴ + 3/x⁵

For the induction one, for the third step, use n = k+2 instead of n = k+1, where the starting term is 2, then continue on as normal.

 

[STx]

you cant do that since there is no common term, the x is only there in x^2 and x , the 3 is x-less
and thanks alot for your help there mate
those two questions stumped me like hell :S worked on them for ages and zit. nada =\

Why cant you do that?

∫(x²+x+3)/3x⁵ dx
=(1/3)∫(1/x³)+(1/x⁴)+(3/x⁵)dx
.
.
.

 

[tehsky]

Why cant you do that?

∫(x²+x+3)/3x⁵ dx
=(1/3)∫(1/x³)+(1/x⁴)+(3/x⁵)dx
.
.
.

wat on earth did u do thurr? ;o

 

[SoulSearcher]

He split up the fraction into individual parts.

You can see that 1/3 * (1/x³+1/x⁴+3/x⁵) = (x²+x+3)/3x⁵ if you add up the LHS.