NewStats: 3,263,591 , 8,180,673 topics. Date: Friday, 06 June 2025 at 05:17 PM 3z4o3k

Allenkee: Please help me on this
On an air Indian flight, there are 9 boys, 5 Indian children, 9 men, 7 Foreign boys, 14 Indian, 6 Indian males and 7 foreign females what is the number of people in the Plane?
Please help.

2 Indian boys + 4 Indian men + 7 foreign boys + 7 foreign females + 3 Indian girls + 5 Indian women + 5 foreign men = 33 people

Total number of people in the plane is 33

To : the question says...

9 boys; which are: 2 Indian boys + 7 foreign boys
5 Indian children; which are: 2 Indian boys + 3 Indian girls
9 men; which are: 4 Indian men + 5 foreign men
7 Foreign boys;
14 Indian; which are: 2 Indian boys + 4 Indian men + 3 Indian girls + 5 Indian women
6 Indian males; which are: 2 Indian boys + 4 Indian men
7 foreign females



Computer scientists or anyone that can help, pls help with this assignment: write a program in visual basic to calculate the compound interest payable in a for any number of years using a maximum of five tools or objects. Thanks.

I'm a biochemistry student - 200 level, but computer science is compulsory for me

Also, I would appreciate tutorials on Visual Basic Programming

Computer scientists, pls help with this assignment: write a program in visual basic to calculate the compound interest payable in a for any number of years using a maximum of five tools or objects. Thanks.

I'm a biochemistry student - 200 level, but computer science is compulsory for me

Also, I would appreciate tutorials on Visual Basic Programming

Christine01: @ Daystarn, its molecular cell biology.I want to combine active research work with lecturing.

I have some e-books on Biochemistry:

Garret and Grisham- Biochemistry 2nd edition.
Colour atlas of Biochemistry, 2nd edition.
Biochemistry, 3rd edition by Mattews, Van Holde and Ahern.
Oxford Dictionary of Biochemistry and molecular Biology.
Biochemistry, 5th edition- Lubert Stryer
Principles of Biochemistry- Lehninger
Harper's Biochemistry, 26th edition... And some others.

Thanks!!! I was able to @ bolded with the names u wrote. I already av @ italised. Do u mind sending d remaining to my e-mail? (Pls check my signature 4 d e-mail)
Thanks.

I'm a prospective 200 level BCH student.

reedbam: Yeah it is possible in almost all federal universities so far your latter uni and course are not having accreditation problem and has a good standing with NUC. That bn said, the process is very rigorous right from your departmen, a letter to your hod-to dean-to and then Dvc academics which they will all minute and make correspondence too. Then you'll need recommendations, etc.. see bro you won't survive it , Infact your hod will tell you blatantly 'I don't have time for this'

That's just your school ohh, I think to get in unilorin is a bit easy but you'll pay abt 400k* (the form is on the site) UDUS too. I have the pdf from Nuc about how to go about but I can't find it, so advise you just chill and pray lasu will be stable. The process is he.ll

Thanks
I think now I'll have to manage

HARDUBUY:
make enquires about unilorin, cos they accept transfer only if they have ur dpartment.

Thanks a lot

Thank you all

ismokeweed: Inter-university transfer was stopped years ago. Long and short, you can not transfer ceteris paribus. But this is Nigeria. Maybe If the VC of UI is your boy.

Thanks

Banjo usman: Go and write jamb again nah, if u re tired or frustrated, am also a product of LASU a final yr student to be precise, just know that it's not happening in ur school only, some other schools are also having issues with the government and lecturers, just pray to God to resolve all dis crises in school, shikena

LASU own is the worst, every aspect of LASU has one issue or the other. I can't write jamb again

Fash20: 8ut why do u want to leave

LASU has made me psychologically weak, currently - formerly I can finish a topic or two within a day, but now I hardly finish one within a week, due to too much sleep, play, sometimes I'll just stay idle.
God help us. Amen

FrancisTony: No.


Tag- Divepen, Taylor184, Delightful1, leystra etc

Thanks

Pls, is it possible for a student of Biochemistry (about to start 200 level, on 1st class, 4.70 cgpa - 5.00 grading system) studying in Lagos State University (LASU) transfer to University of Ibadan (UI)? If possible, how?
Note: the student will start with the current prospective 200 level students at UI.
Thanks in advance.

Pls, is it possible for a student of Biochemistry (about to start 200 level, on 1st class, 4.70 cgpa - 5.00 grading system) studying in Lagos State University (LASU) transfer to University of Ibadan (UI)? If possible, how?
Note: the student will start with the current prospective 200 level students at UI.

Contributions, suggestions and comments are welcome. Thanks in advance.

...Still ugly.

Laplacian:
i thought u 've resolved it:
now,
f(x,z)=f(x)+zf'(x)+z²f"(x)/2!+...

If f(x,z)=tan^-1(x+z), then the above equation resolves into
tan^-1(x+z)=tan^-1x+zf'(x)+z²f"(x)/2!+...
Since f(x)=tan^-1x
then,
f'(x)=1/(1+x²)

f"(x)=-2x/(1+x²)²=-2x[f('x)]²
f'"(x)=d(-2x[f('x)]²)/dx, e.t.c
now since cot(z)=x, => cot²(z)=x², or 1+x²=1+cot²z=cosec²x,
:. sin²x=1/(1+x²),
i hop u can complet the rest

*i must comment that, your series is slightly INCORRECT*

Thank you very much Sir

benbuks:

sorry i could‘nt. solve ur question.....

did‘nt av tym to look into it well....

mayb i dont understand it well..
al d same ..our able gurus will do justice....tnx 4ur understandin..

1love

All is cool

akpos4uall: ^^^
Instead of those plenty plenty wahalai above for problem 3, we can just state that the (r + 1)th term in the expression (1 + x)ⁿ is ⁿC_r * x^r
Where r = 0, 1, 2, ... n
ⁿC_r = n!/(r! * (n - r)!)
:. The (r + 1)th term of x²ⁿ(1 + x^-2)²ⁿ = ²ⁿC_r * x^-2r * x²ⁿ
= ²ⁿC_r * x^{2n - 2r}
From here we can arrive at the right answer as posted earlier.

Pls help me with this

Prove that tan^-1(x+z) = tan^-1(x) + 2sin(z)sin(z) - [(2sin(z))^2sin^2(z)]/2 + [(2sin(z)^3sin^3(z)]/3 - ...
Where cot(z) = x
Or clearer,
Let 2sin(z) = y
We now have
Prove that tan^-1(x+z) = tan^-1(x) + ysin(z) - (y^2/2)sin^2(z) + (y^3/3)sin^3(z) - (y^4/4)sin^4(z) + ...
Where cot(z) = x
(Question under TAYLOR'S EXPANSION)

Thanks

akpos4uall:

Let me give it a try
(1 + x)ⁿ = 1 + nx + n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + n(n - 1)(n - 2)(n - 3)x⁴
+ ...
This can be written as
(1 + x)ⁿ - 1 - nx = n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + n(n - 1)(n - 2)(n - 3)x⁴ + ... Eqn (1)

Let z = 5/(3 * 6) + 5 * 7/(3 * 6 * 9) + 5 * 7 * 9/(3 * 6 * 9 * 12) + ...
Multiply both sides by 3

3z = 3 * 5/(3 * 6) + 3 * 5 * 7/(3 * 6 * 9) + 3 * 5 * 7 * 9/(3 * 6 * 9 * 12) + ...
3z = 3 * 5 * (1/3)²/(1 * 2) + 3 * 5 * 7 * (1/3)³/(1 * 2 * 3) + 3 * 5 * 7 * 9 * (1/3)⁴/(1 * 2 * 3 * 4) + ...

3z = 3 * 5 * (1/3)²/2! + 3 * 5 * 7 * (1/3)³/3! + 3 * 5 * 7 * 9 * (1/3)⁴/4! + ... Eqn (2)

Comparing eqn (1) & eqn (2)
3z = (1 + x)ⁿ - 1 - nx Eqn (3)

3 * 5 * (1/3)² = n(n - 1)x²
5 = 3n(n - 1)x² Eqn (4)

3 * 5 * 7 * (1/3)³ = n(n - 1)(n - 2)x³
35 = 9n(n - 1)(n - 2)x³ Eqn (5)

3 * 5 * 7 * 9 * (1/3)⁴ = n(n - 1)(n - 2)(n - 3)x⁴
35 = 3n(n - 1)(n - 2)(n - 3)x⁴ Eqn (6)

Divide equation (5) by equation (4)
7 = 3x(n - 2) Eqn (7)

Divide equation (6) by equation (5)
1 = x(n - 3)/3
9 =3x(n - 3) Eqn (

Divide equation ( by equation (7)
9/7 = (n - 3)/(n -2)
9(n - 2) = 7(n - 3)
9n - 18 = 7n - 21
2n = -3
n = -3/2
Substitute -1.5 for n in equation (
9 = 3x(-1.5 - 3)
9 = 3x * -4.5
x = 9/(3 * -4.5)
x = -2/3 , n = -3/2

z is what we are looking for. Hence substitute the obtained values of n & x into equation (3)
3z = (1 + x)ⁿ - 1 - nx Eqn (3)
3z = (1 + -2/3)^(-3/2) - 1 - (-2/3)(-3/2)
3z = (1/3)^(-3/2) - 1 - 1
3z = 3^(3/2) - 2
3z = (3^3)^(1/2) - 2
3z = 27^(1/2) - 2
z = 3 * 3^(1/2)/3 - 2/3
z = 3^(1/2) - 2/3

Cool approach there

Thanks a lot

akpos4uall:

The answer is -2/3 + sqrt 3
The process just too long

Pls sir, try and post the solution for us

benbuks: ^ ok...m on bed nw..mayb i ‘d try dem 2maro...quit cheep sha.

I'm still waiting for ur solutions Sir

Prove that tan^-1(x+z) = tan^-1(x) + 2sin(z)sin(z) - [(2sin(z))^2sin^2(z)]/2 + [(2sin(z)^3sin^3(z)]/3 - ...
Where cot(z) = x

It now remains the solution for this last question, but u can still post ur solutions 4 d oda questions if they are different from d ones posted

Thanks

Laplacian:
k, we know that, for a series to be representable in a binomial form, the must have a recurrent structure, and the numerator & denominator must have the same number of factors. The two conditions are satisfied for case u solved, since
T_n=#(2r+1)/[4^n-1(n-1)!]=T_n-1*(2n+1)/[4^n-1(n-1)!]
the one i solved also to satisfies:
T_n=#(2r+3)/[3ⁿ⁺¹*(n+1)!]=T_n-1(2n+3)/[3ⁿ⁺¹*(n+1)!], but it fails for the nunber of factors, to accommodate that, consider the eqn;
(1+x)ⁿ=1+nx+[n(n-1)/2!]x²+[n(n-1)(n-2)/3!]x³+...
I think a little modification can make it suit the conditions
e.g
(1+x)ⁿ-(1+nx)=[n(n-1)/2!]x²+[n(n-1)(n-2)/3!]x³+...
We divide thru by nx to obtain
[(1+x)ⁿ-(1+nx)]/nx=[(n-1)/2!]x+[(n-1)(n-2)/3!]x²+[(n-1)(n-2)(n-3)/4!]x³+...
Use ur method for the last eqn lets c hw far it goes...

I tried it but couldn't solve further to arrive at an answer; let's leave the answer at: Tn=[(3n+2)!/3!]/(2n/2•[(3n+2)/2]!)/[3n+1•(n+1)!]
Where u stopped initially

Thanks a lot, I really appreciate

Laplacian:
yea, i made the error while typin, u postd ur comment while i was editin mine, its already correctd.
nice one!
So u 're in possession of d solution afterall

D one I solve is different from d one u solved, and its a bit easier.
Thanks man

Laplacian:
for ur no.1, am not really sure what u mean by "odd and even ", but here's a suggestion;
1.) expansion of (x+a)^b should contain only "odd & even ", so that, from the condition of the question,
a+b=(x+a)^b, then multiplying thru by a-b, we obtain;
a²-b²=(x+a)^b•(a-b)..

2.) this one is quiet 'cheap', given
f(x)=
(1 - x + x²)⁴. Take the fouth order derivative of the above expression, set x=0, and divide the result by 4!...
Simply put, obtain f^iv(0)/4!
In case u ain't satisfied with that, here is an elementary
ALTERNATIVE SOLUTION
(1 - x + x²)⁴=[1+(x-1)x]⁴=1+4(x-1)x+6[(x-1)x]²+4[(x-1)x]³+[(x-1)x]⁴, inspection shows that only the last three contains x⁴, and the coefficient can be readily obtained from there.
3.) let # denote pi, the product of an expression in d variable r, from r=1 to n, by inspection, the numerator of each term of ur series is the product of an A.P with fomula; #(3r+2), similar, the denominator is the product of a G.P with the formula; 3ⁿ⁺¹•(n+1)!, so the n^th term of ur sequence is;
T_n=[#(3r+2)]/[3ⁿ⁺¹•(n+1)!]
Recall that,
#(3r+2)=5•7•9•11•13•15...
=#[(3r+1)•(3r+2)]/#(3r+1)
=(4•5•6•7....)/(4•6•8•10•12...)
(i proceed for the case in which n is even, use n+1 for the index of 2 if n is odd)
#(3r+2)=

[(3n+2)!/3!]/(2ⁿ•3•4•5•6...)
=[(3n+2)!/3!]/(2ⁿ/2•[(3n+2)/2]!)
so that,
T_n=[(3n+2)!/3!]/(2ⁿ/2•[(3n+2)/2]!)/[3ⁿ⁺¹•(n+1)!]

the solution becomes too complicatd to paste here afterward...i'll paste the final solution soon...

I'm clear with d no 1 and 2 solutions, but not very ok wit d 3rd - d numerator series formula
D formula I hav for d numerator is #(5+(n-1)2)

Have a look at dis:

Sum the series
1 + 3/4 + 3•5/4•8 + 3•5•7/4•8•12 + ...

SOLUTION

(1+x)^n = 1 + nx + [n(n-1)/2!]x^2 + [n(n-1)(n-2)/3!]x^3 + ... + x^n
We find nx = 3/4 ...........(1)
[n(n-1)/2!]x^2 = 3•5/4•8 .........(2)
From (1)
x = 3/4n .......(3)
Putting (3) into (2)

[n(n-1)/2](3/4n)^2 = 3•5/4•8
9n(n-1)/32n^2 = 15/32
3(n-1)/n = 5
(1-1/n) = 5/3
1/n = 1-5/3
1/n = -2/3

n = -3/2 ......................

From x = 3/4n = 3/4(1/n) = 3/4(-2/3)

x = -1/2 ......................

Hence the series is (1 + x)^n = (1 - 1/2)^(-3/2) = 1 + 3/4 + 3•5/4•8 + 3•5•7/4•8•12 + ...
Solving (1 - 1/2)^(-3/2) gives 2(2)^(1/2) meaning 2 * square root of 2

Thanks a lot bro

doubleDx:

^

Nice one, though the problem has already been solved by AYSUCCESS99. Why not finish the rest if you have the solutions? The reason why solutions to some questions are not posted quickly is because typing them out isn't all that easy, it's not because your questions are too difficult or "hard" for gurus here to solve.

For the benefit of everyone, if anyone has a solution to his/her problem, it's better they post them if the questions are not attended to quickly; that way, others would learn from from them, instead recycling the problems over and over again! I'm saying this because a lot of people post questions and leave them unsolved if the the clinic's active do not make efforts to attempt them. We are building this all thing to encourage mathematics lovers of today and future leaders, so let's do it with all our hearts and might; it's not about competition!

Thank y'all and happy new year!

Aysuccess99 used normal expansion which is tedious - not a good idea in exam situations, but still got it correctly. I used binomial expansion which is easier; I actually solved it TODAY.

As for the other questions, I am yet to solve them that's why I posted all 4 questions.

ONE LOVE

The coefficient of x^4 in the expansion of (1 - x + x^2)^4 is __

SOLUTION

Let: y = 1 - x
(y + x^2)^4
Expanding by binomial expansion yields
y^4 + 4y^3x^2 + 6y^2x^4 + 4yx^6 + x^8
From the above:
y^4 + 4y^3x^2 + 6y^2x^4 gives a coefficient of x^4

Separately, we have
y^4 = (1 - x)^4 = 1 - 4x + 6x^2 - 4x^3 + x^4
We have 1 as coefficient of x^4 here

4y^3x^2 = 4[(1 - x)^3x^2] = 4[(1 - 3x + 3x^2 - x^3)x^2] = 4(x^2 - 3x^3 + 3x^4 - x^5) = 4x^2 - 12x^3 + 12x^4 - 4x^5
We have 12 as coefficient of x^4 here

6y^2x^4 = 6[(1 - x)^2x^4] = 6[(1 - 2x + x^2)x^4] = 6x^4 - 12x^5 + 6x^6
We have 6 as coefficient of x^4 here

Adding all coefficients of x^4 yields: 1 + 12 + 6 = 19
Hence, coefficient of x^4 = 19


Thanks again @aysuccess99

Laplacian:
if u tink they are cheap, den we 'll patiently wait 4 ur solution...

Happy new year to u all

benbuks: ^ ok...m on bed nw..mayb i ‘d try dem 2maro...quit cheep sha.

benbuks: ..ok..make them clearer n solvable

Ok let's say 2sin(z) = y
We now have
tan^-1(x+z) = tan^-1(x) + ysin(z) - (y^2/2)sin^2(z) + (y^3/3)sin^3(z) - (y^4/4)sin^4(z) + ...
Where cot(z) = x
(Question under TAYLOR'S EXPANSION)


Would hav snapped dem long b4 now, but 4 ASUU strike - I didn't fully comprehend SIMPLE SERIES b4 d strike interrupted.

D 1st and 2nd questions are from BIONOMIAL THEOREM/EXPANSION

D 3rd is under APPLICATION OF BINOMIAL THOREM

benbuks: ..snap the last question.n post..

I'm yet to do dat bro...
and it's actually 4 questions

jackpot: Clear.


Congrats on resolving your 3-yrs old question. See the power of sharing!!!

Any other enigmas for us to brainstorm on?

Hmm, doc of nairaland maths clinic, solve my 4 questions na

If 'a' is the sum of odd and 'b' the sum of even in the expansion (x + a)^b , then find a^2 - b^2

The coefficient of x^4 in the expansion of (1 - x + x^2)^4 is __

Sum the series:
[5/(3 • 6)] + [(5 • 7)/(3 • 6 • 9)] + [(5 • 7 • 9)/(3 • 6 • 9 • 12)] + ...~
NB: the • means multiplication as in, 2 • 3 = 6

Prove that tan^-1(x+z) = tan^-1(x) + 2sin(z)sin(z) - [(2sin(z))^2sin^2(z)]/2 + [(2sin(z)^3sin^3(z)]/3 - ...
Where cot(z) = x

Thanks in anticipation...

Seriously now, I'm cool wit u

doubleDx:

Lol, she is actually one of the best in the clinic! I will attempt your questions when I'm free. 1luv and no beefs bruv....You are welcome!

Thanks bruv, it's cool wit me now

I'm still waiting for solutions to my questions: check d previous page, or maybe @JACKPOT will help me wit dem