Swagalord18: prove that if the first term of an A.P is 1, and the sum of n terms is n², the A.P is 1,3,5.....9

the formula for the sum of n terms of an AP is

Sn=ⁿ/₂[2a+(n-1)d]
where a is the first term and d is the common difference.
Therefore by putting a=1(given), we have
Sn=ⁿ/₂[2+(n-1)d]

Equate this sum to n²(given) and simplify(since n not equal to 0), to get
n= ¹/₂[2+(n-1)d]
therefore
2n=2+(n-1)d
2n=2+dn-d
2n-dn=2-d
n(2-d)=2-d
n(2-d)-(2-d)=0
(n-1)(d-2)=0
from which we have
d-2=0
d=2

first term a=1 (given)
2nd term =a+d=1+2=3
3rd term=a+2d=1+2(2)=5
and so on.

Thats all!

Hi doubleDx

what's the name and author of that textbook you uploaded while solving that paraboloidal stuff?

Swagalord18: prove that if the first term of an A.P is 1, and the sum of n terms is n², the A.P is 1,3,5.....9

here is it...
Take T(n)=T subscript nd S(n),S(n-1)=s sub n&S sub (n-1) respectively
generally T(n)=S(n)-S(n-1)...an establishd fact
also if S(n)=n^2
:- S(n-1)=(n-1)^2
hence d above gvs
T(n)=[n^2]-[n-1]^2
when n=1(u can chk any chosen no)
substn we have
T(1)=1-0=1
T(5)=25-16=9...hope it helps

jackpot:
the formula for the sum of n terms of an AP is

Sn=ⁿ/₂[2a+(n-1)d]
where a is the first term and d is the common difference.
Therefore by putting a=1(given), we have
Sn=ⁿ/₂[2+(n-1)d]

Equate this sum to n²(given) and simplify(since n not equal to 0), to get
n= ¹/₂[2+(n-1)d]
therefore
2n=2+(n-1)d
2n=2+dn-d
2n-dn=2-d
n(2-d)=2-d
n(2-d)-(2-d)=0
(n-1)(d-2)=0
from which we have
d-2=0
d=2

first term a=1 (given)
2nd term =a+d=1+2=3
3rd term=a+2d=1+2(2)=5
and so on.

Thats all!

....nice one...2 dat ur quest abt the set... she is currently in her finals..maths dpartmnt ...guess itz nt u ryt?

If we drop a stone,we can assume air resistance ("drag") to be negligible. Experiments show that under that assumption the acceleration y"= d^2 y/dt^2 of this motion is constant (equal to the so-called acceleration of gravity g=9.80 ms^-2 =32ftsec^-2) . State this as an ODE for y(t), the distance fallen as a function of time. Solve the ODE to get the familiar law of free fall, y=gt^2 / 2

benbuks: If we drop a stone,we can assume air resistance ("drag" to be negligible. Experiments show that under that assumption the acceleration y"= d^2 y/dt^2 of this motion is constant (equal to the so-called acceleration of gravity g=9.80 ms^-2 =32ftsec^-2) . State this as an ODE for y(t), the distance fallen as a function of time. Solve the ODE to get the familiar law of free fall, y=gt^2 / 2

na to just dey integrate, substitute, and apply initial conditions.

It is not challenging enough

2nioshine:
here is it...
Take T(n)=T subscript nd S(n),S(n-1)=s sub n&S sub (n-1) respectively
generally T(n)=S(n)-S(n-1)...an establishd fact
also if S(n)=n^2
:- S(n-1)=(n-1)^2
hence d above gvs
T(n)=[n^2]-[n-1]^2
when n=1(u can chk any chosen no)
substn we have
T(1)=1-0=1
T(5)=25-16=9...hope it helps

smarter solution.

Guess not all will understand the step sha, especially the fact that Tn=Sn-(Sn-1).

With this method, the formula for the nth term is gotten "free"

jackpot: na to just dey integrate, substitute, and apply initial conditions.

It is not challenging enough

..do, than say.

2nioshine: ....nice one...2 dat ur quest abt the set... she is currently in her finals..maths dpartmnt ...guess itz nt u ryt?

nah.

benbuks: ..do, than say.

integrate first, you get
y'=gt+c₁
y' is velocity and at t=0, the initial velocity y'(0) is 0 since the particle is falling from rest. Substitute these facts to get c₁=0.
Thus,
y'=gt.
Integrate again to get
y=gt^2+c₂

but at t=0, the distance covered is zero. Substitute these into the above equation to get c₂=0.

Thus,

y= gt^2/2

thats all!

rhydex 247: Here is my question.
1. Solve for x. If x+a^(x^a)=b. Where x=(x1,x2,x3). Note that ^ means cap.

2. Let W be a subspace of real space R^3. Give a geometrical describtion of W in terms of its dimension.

i ve very high esteem 4 ur questns
If x+a^(x^a)=b, from d structure of x we can infer dat d above eqn is a vector eqn, applyin vector tripl produt to d 2nd term on d L.H.S,
x+x(a.a)-a(a.x)=b....eqn1, multiply d given eqn by a. to get,
a.x+a.[a^(x^a)]=a.b, scalar tripl prodct givs
a.x-(x^a).[a^a]=a.b, recall dat
a^a=0, so a.x=a.b, substitt in eqn1, to get; x(1+|a|^2)-a(a.b)=b, so
x=[b+a(a.b)]/(1+|a|^2)

Alpha Maximus: ,..For real? Well I guess you ARE serious. Let's just get this over with!!!
3 7/8 +1 1/3 =5 5/24
1 2/3 -3/8 =1 7/24
This question is invalid seeing as the first result is clearly greater than the second, so it isn't less than the difference of the aforementioned fractions by anything, at least not positive

Alpha Maximus: ,..For real? Well I guess you ARE serious. Let's just get this over with!!!
3 7/8 +1 1/3 =5 5/24
1 2/3 -3/8 =1 7/24
This question is invalid seeing as the first result is clearly greater than the second, so it isn't less than the difference of the aforementioned fractions by anything, at least not positive

am greatful that @least someone attempt tha question.
This question is just like one of those question being given during work interview especially if u are a student.
It actually looks simple but then men have fallen on a mission to solving it
tha answer to tha question is
3 2/3.

Please house, help me solve these questions with explanation:

(1) Convert 123.53 base 10 to base 3
(2)0.36 base 10 to base 8
(3) 23.123 base 10 to base 5

Note: I did it by dividing the two parts by the required bases but the answers are different. I'm pussled. Help me out.

Hmmmm

jackpot:
the formula for the sum of n terms of an AP is

Sn=ⁿ/₂[2a+(n-1)d]
where a is the first term and d is the common difference.
Therefore by putting a=1(given), we have
Sn=ⁿ/₂[2+(n-1)d]

Equate this sum to n²(given) and simplify(since n not equal to 0), to get
n= ¹/₂[2+(n-1)d]
therefore
2n=2+(n-1)d
2n=2+dn-d
2n-dn=2-d
n(2-d)=2-d
n(2-d)-(2-d)=0
(n-1)(d-2)=0
from which we have
d-2=0
d=2

first term a=1 (given)
2nd term =a+d=1+2=3
3rd term=a+2d=1+2(2)=5
and so on.

Thats all!

ur very good ooh ...kudos
nd i lyk d way u take ur tym 2 make use of the sub nd sup 4 beta understanding.
.
Take this as desert
.
The nth term of the series 5 + 5 + 6¹/₂ + ..... Is given by T_n = a(¹/₃)^n-1 + bn. Find the values of a and b, and the sum of the first n terms of the series

jackpot: integrate first, you get
y'=gt+c₁
y' is velocity and at t=0, the initial velocity y'(0) is 0 since the particle is falling from rest. Substitute these facts to get c₁=0.
Thus,
y'=gt.
Integrate again to get
y=gt^2+c₂

but at t=0, the distance covered is zero. Substitute these into the above equation to get c₂=0.

Thus,

y= gt^2/2

thats all!

,,jacpot jackpot jackpot...ooo jackpot my love.;-).y did u choose 2 break ma heart....o jackpot..avta all av done 4u.











Any sha i remove my slippers 4u....just wanted odas 2 learn.

benbuks: ...mayb..am

i c.....

To whom brain is given, mathematics is expected.

wisemania:
i c.....

....r u...?

jackpot: integrate first, you get
y'=gt+c₁
y' is velocity and at t=0, the initial velocity y'(0) is 0 since the particle is falling from rest. Substitute these facts to get c₁=0.
Thus,
y'=gt.
Integrate again to get
y=gt^2+c₂

but at t=0, the distance covered is zero. Substitute these into the above equation to get c₂=0.

Thus,

y= gt^2/2

thats all!

mth125....fym u be super star...kip it up,let me go n sip some shekpe and some dogo b4 i strt my own display.....

benbuks: ....r u...?

45p1r1n9.....

wisemania:


45p1r1n9.....

9006

Swagalord18:
ur very good ooh ...kudos
nd i lyk d way u take ur tym 2 make use of the sub nd sup 4 beta understanding.
.
Take this as desert
.
The nth term of the series 5 + 5 + 6¹/₂ + ..... Is given by T_n = a(¹/₃)^n-1 + bn. Find the values of a and b, and the sum of the first n terms of the series

Hey bro, check this, from where you copied it out. It should be 6¹/₃.

Having said that, put n=1 and n=2 in the T_n equation and equate to 5 and 5 respectively.

You will then get the two equations

a+b=5 . . . . . . . . . . . . . . . . . . . (1)
^a/₃+2b=5. . . . . . . . . . . . . . . . .(2)

solve these simultaneously to get
a=3, b=2

substitute these in the T_k equation(I am calling it T_k instead of T_n because I want to sum from k=1 to n. Remember they are dummy variables after all )

T_k= 3(¹/₃)^k-1 + 2k

sum over all n and simplify (you should use the sum of a GP formula on the first summand and the sum of an AP formula on the second summand here).

The tiny details are left as an easy exercise for the reader

Final answer is

S_n = ⁹/₂ [1-(¹/₃)ⁿ ] + n(n+1)


Thats all!

benbuks: 9006

bro am tired of seeing u guys solve problems here evryday...maths is not all bout theory,der shud be som practical aproach 2 it...i posted an idea of d application of maths in AI(artificial intelligence),how 2 code it..using 3D maths and physics,but its like no 1 saw wt i was seeing,dey r all interested in d theoretical aspect of maths....am not happy about dat..d wrld is undergoin sponteneous changes,we nid 2 change along wit it....lets invest all dis mths knowledge in2 3D modelling naw,u guys av all it takes.....try n make some research on it...and also visit d programming section of this forum....na beg i dey beg una...ejoor

wisemania:
mth125....fym u be super star...kip it up,let me go n sip some shekpe and some dogo b4 i strt my own display.....

hey bro, no blame me much for solving that small question sha.

I am acting on instructions by my 5-STAR GENERALS sha. I solve the moderate ones. I refer only the tougher questions to them. They are the best.

busuyem: Please house, help me solve these questions with explanation:

(1) Convert 123.53 base 10 to base 3
(2)0.36 base 10 to base 8
(3) 23.123 base 10 to base 5

Note: I did it by dividing the two parts by the required bases but the answers are different. I'm pussled. Help me out.

i can't precisely remember d method i was taught in school, but here's one i just deviced, it may help....i know where ur problem lies-d decimal part-follow d algorithm below;
1)convert d decimal part to fraction; e.g123.53 to base3...
(123=11120base3), 0.53=53/100
2) multiply d fraction by d base u are convertin to;
3*53/100=159/100=1+59/100
3) keep d quotient (1)
4) multiply d remainder/divisor by d base u are convertin to;
3*59/100=177/100=1+77/100
5)keep d quotient (1)
6)multiply d remainder/divisor by d base u are convertin to;
3*77/100=231/100=2+31/100
7) keep d quotient(2)
multiply d remainder/divisor by d base u are convertin to;
3*31/100=93/100=0+93/100
9) multiply d remainder/divisor by d base u are convertin to;
3*93/100=279/100=2+79/100
10)keep d quotient......
D process continues like dat indefinitely.....now all d quotients u kept, in descending order(11202), is d required conversion to d given base(3)......
123.53base10=11120.11202base3........all other conversions can be done similarly....i have d proof but its not necessary....a table is necessary for clarity, ask JACKPOT to make a table for u, her device is a bit powerful....

wisemania:
bro am tired of seeing u guys solve problems here evryday...maths is not all bout theory,der shud be som practical aproach 2 it...i posted an idea of d application of maths in AI(artificial intelligence),how 2 code it..using 3D maths and physics,but its like no 1 saw wt i was seeing,dey r all interested in d theoretical aspect of maths....am not happy about dat..d wrld is undergoin sponteneous changes,we nid 2 change along wit it....lets invest all dis mths knowledge in2 3D modelling naw,u guys av all it takes.....try n make some research on it...and also visit d programming section of this forum....na beg i dey beg una...ejoor

..yea bruv. av heard u...am nt 2 good in progaming though,..i will c what i cn do abt dat....mayb u shud gv me d application u r using..

Laplacian:
i can't precisely remember d method i was taught in school, but here's one i just deviced, it may help....i know where ur problem lies-d decimal part-follow d algorithm below;
1)convert d decimal part to fraction; e.g123.53 to base3...
(123=11120base3), 0.53=53/100
2) multiply d fraction by d base u are convertin to;
3*53/100=159/100=1+59/100
3) keep d quotient (1)
4) multiply d remainder/divisor by d base u are convertin to;
3*59/100=177/100=1+77/100
5)keep d quotient (1)
6)multiply d remainder/divisor by d base u are convertin to;
3*77/100=231/100=2+31/100
7) keep d quotient(2)
multiply d remainder/divisor by d base u are convertin to;
3*31/100=93/100=0+93/100
9) multiply d remainder/divisor by d base u are convertin to;
3*93/100=279/100=2+79/100
10)keep d quotient......
D process continues like dat indefinitely.....now all d quotients u kept, in descending order(11202), is d required conversion to d given base(3)......
123.53base10=11120.11202base3........all other conversions can be done similarly....i have d proof but its not necessary....a table is necessary for clarity, ask JACKPOT to make a table for u, her device is a bit powerful....

I really appreciate this but the question is, how do I knw the terminating point when multiplying by the base? How do I knw where to stop multiplying? Help me out pls.

busuyem:


I really appreciate this but the question is, how do I knw the terminating point when multiplying by the base? How do I knw where to stop multiplying? Help me out pls.

,...number base conversions involving decimals dont always terminate..so u stop at 4 decimal place..

...this reminds me. In my school,a professor of computational mathematics delivered a lecture on "ANOMALIES IN HARMONING OF MATHEMATICAL PRECISION". And in that lecture,he mentioned some anomalies in mathematics we often use in our day to day calculation from local to professional levels,e.g pi=22/7~3.14159.. But we mostly use 3.142(first anomaly) also e^x=1+x/1! + x^2/2! + x^3 /3! +..., (-infinity < x< infinity..).when x=05 for .e^x ~ 1 =>e^0.5~ 1
.e^x ~ 1+ x/1! =>e^0.5~ 1.5
.e^x ~ 1+x/1! +x^2/2! =>e^0.5~ 1.625
.e^x ~ 1 +x/1!+ x^2/2! +x^3/3! =>e^0.5~ 1.624833333
. And so on....[anomaly due to truncation of taylor series expansion]....there are lots more of anomalies in mathematics,though we cant totally eliminate them, but we could at least reduce their frequcies....the more number of decimal places you use in your calculation, the more accurate your solution and final results will tend.

True yarn broda....

benbuks: ..yea bruv. av heard u...am nt 2 good in progaming though,..i will c what i cn do abt dat....mayb u shud gv me d application u r using..

well said....actually am not in2 2D or 3D modelling/animation for now coz am stil developin my webdesign skills...but in d tym beyond i might delve in2 animation tinz...3D maths and fiziks dey scare me sha,i neva get dia tym 4now sha....make una dey solve maths dey go,i go juz dey 4 korna dey observe una dadaly and wellemly....