chizzy94:
Memories are Everything .

Neldrizzy:
INTERNATIONAL STUDIES and DIPLOMACY (ISD) aspirant like ♍ε̲̣̣̣, hallaaaaaaaa #neldrizzy

nobody:
Me just dey my own dey observe .Neldrizzy u still dey play war scatter?

Emperorking20:
Kulkatty, 3coins, Efedua, Suley316, benromeo39, Chizzy94, pweetyz, Neldrizzy, bestlily, fortunate, Victorious1, CJPresh, ahika,Nimen, HonourablePomk, Mesoade, chue, incredyable, and all other unmentioned team Uniben.. I'm grateful

Naijaboiy, tosyne2much, Amefrica, Kiddablingzz, Chrisviral, Rilwayne001, Chibwike, and all other unmentioned contestants.. I'm grateful

Kelechimarie, chimkaire, midasbliss, marieolae, PinkObsession, Epaul, immortalvoices, my boss Cwaya , tintingz, Jboy, Ercross, tiffae, vivaciousvivi, Victoriousvic, wincyhov, ionsman, leki10, emyben, missmossy, sugarcoatted, missterious, GRIMMJOE, luckygurl, olagockay, muguliciouMUGU, my honourable Macklef , 6packdude, homesteady, striktlymi, abpatrick7, VICTOROCHEI, VICTORCIZA, izinhopro, xtarxhyne, ObiChiz08, gabialonso, my cool man danaik, roufy235, destante, mirexxx, Nursy, jalie and everyother unmentioned pal.. I'm grateful

It was your support that brought me here today, it is that which has kept me in ds race, it is what will still keep me going, it is your support and votes, and I'm thankful..

I'm thankful for ur time, ur mb , ur persistent efforts and that's just to say d least..

What pains me to no end is d fact that "Thank u!" Is d only thing I can say and offer, what also gives me joy is d fact that u'll accept my thanks

Oh! How I wish I could do better, sincerely!
Thanks a lot guys.. Slowly but surely, d lot of u have contributed in a meaningful way to my life, slowly but surely, u guys are becoming persons I can remember with a smile playing on my lips, slowly but surely u guys have become my PALS

I'm saying a big thank u to y'll! Thank u very much for yesterday, today and d days to come..
In or out, I don't care! I'm grateful and I owe u lot one
Thank u very much!!!

With love!!
Emperorking20

RileyFreeman:


what!!!!! + + = 3



My brother, Good thing you visited this thread before purchasing the form.

The subjects are;

1. English Language.
2. Government/ History
3. C.R.S
4. literature in English

somebody:
eya see as this poor boy dey lament him no admission status but shall I no shake God dey, said Mr Johnson, pls med/surgery aspirant add me on whatsapp

Suley316:
... Seems we are goin the same way

efedua:
let n be the numbers of people placed on line to have a shakes with each others. obviously, the highest possible count of shakes,will be from the first person and must be given as 'n-1=a' and the lowest possible will be given as 'a-(n-2)=1'
however when we run a summation of shakes from nth term(being the first person shakes to others and left) to 1th term(being the last single person who will not shake any body and left) or a to 0 we will have an AP of a+a-1+a-2+....+a-(n-3)+a-(n-2)+a-(n-1)
.
(where a-(n-1)=0)
.
when we count the numbers of terms in the summation above we will get n numbers of terms.
prove:
when 3people have shakes, by the summation of a to 0, where n-1=a, we have the numbers of shakes as 2+1+0=3
now count the numbers of terms, we have 2,1 and 0 and they are three terms.

let write the summation from a to 0 as Tn, where Tn is the total numbers of shakes, we get;
Tn=a+a-1+a-2+....+a-(n-3)+a-(n-2)+a-(n-1)
.
Similarly, if we rewrite the summation from 0 to a as Tn*, we get;
Tn*=a-(n-1)+a-(n-2)+a-(n-3)+....+a-2+a-1+a. (where Tn*=Tn, Then Tn +Tn*=2Tn)
.
.
Now let add Tn to Tn* by adding there terms in the same position without affecting the n numbers of terms. that is, add Tn 'a' to Tn* 'a-(n-1)' and Tn 'a-(n-1)' to Tn* 'a' and do the same for all value that fall within them

for the formal we have;
a+a-(n-1)=2a+(n-1)
for the later we also have;
a-(n-1)+a=2a+(n-1)
NB:this constant value 2a+(n-1) will still occur for all added terms by position
prove:
from Tn 2nd term and Tn* 2nd term summation we get;
{a-1}+{a-(n-2)}= 2a-1-(n-2)=2a-1-n+2=2a-n+1=2a-(n-1)
.
.
.
.so therefore, the sum of Tn and Tn*, we give;
2Tn=2a-(n-1)+ 2a-(n-1)+ 2a-(n-1)+....+ 2a-(n-1)+ 2a-(n-1)+2a-(n-1)
By principle of addition; a+a+a+a=4a we have that;
2Tn=n[2a-(n-1)]
©
©
©
©
now substitute 'n-1' for 'a' as in the case. of highest possible shakes. we have;
2Tn=n[2(n-1)-(n-1)]=n[n-1] (NB: 2a-a=a)
.
2Tn=n(n-1)
finally divide L.H.S and R.H.S of the equation by 2 to have Tn, we get;
2Tn/2=[n(n-1)]/2
therefore; Tn=n/2(n-1)

©®
CREDIT to my:
cracking BRAIN

chizzy94:
Prospective Engineering Students whichever one - Chemical, Petroleum, Mechanical, Civil, Production, Structural et al. Should hit 'Like' or 'quote' me. Let see how we get on from here.

You can also follow me on twitter @ASAP_CHizzy. I follow back.
Let us work our way to the University of Benin. Greatest Gbo Gbo!!

chizzy94:

LOL
exactly. We should just laugh at ourselves a little bit. We've done some growing up really.. How time flies.
And we could also push the thread a few pages..
I hail these guys;
3Coins, Suley316, Neldrizzy, Naijaboiy, Idiosyncrasy, JaymeJate et al.. We've been around for so long, lol it's been a long time coming! I couldn't just resist laughing at Drizzy's desperation .
#ThrowBack

efedua:
we dey fine o and u

efedua:
we dey fine o and u

efedua:
let n be the numbers of people placed on line to have a shakes with each others. obviously, the highest possible count of shakes,will be from the first person and must be given as 'n-1=a' and the lowest possible will be given as 'a-(n-2)=1'
however when we run a summation of shakes from nth term(being the first person shakes to others and left) to 1th term(being the last single person who will not shake any body and left) or a to 0 we will have an AP of a+a-1+a-2+....+a-(n-3)+a-(n-2)+a-(n-1)
.
(where a-(n-1)=0)
.
when we count the numbers of terms in the summation above we will get n numbers of terms.
prove:
when 3people have shakes, by the summation of a to 0, where n-1=a, we have the numbers of shakes as 2+1+0=3
now count the numbers of terms, we have 2,1 and 0 and they are three terms.

let write the summation from a to 0 as Tn, where Tn is the total numbers of shakes, we get;
Tn=a+a-1+a-2+....+a-(n-3)+a-(n-2)+a-(n-1)
.
Similarly, if we rewrite the summation from 0 to a as Tn*, we get;
Tn*=a-(n-1)+a-(n-2)+a-(n-3)+....+a-2+a-1+a. (where Tn*=Tn, Then Tn +Tn*=2Tn)
.
.
Now let add Tn to Tn* by adding there terms in the same position without affecting the n numbers of terms. that is, add Tn 'a' to Tn* 'a-(n-1)' and Tn 'a-(n-1)' to Tn* 'a' and do the same for all value that fall within them

for the formal we have;
a+a-(n-1)=2a+(n-1)
for the later we also have;
a-(n-1)+a=2a+(n-1)
NB:this constant value 2a+(n-1) will still occur for all added terms by position
prove:
from Tn 2nd term and Tn* 2nd term summation we get;
{a-1}+{a-(n-2)}= 2a-1-(n-2)=2a-1-n+2=2a-n+1=2a-(n-1)
.
.
.
.so therefore, the sum of Tn and Tn*, we give;
2Tn=2a-(n-1)+ 2a-(n-1)+ 2a-(n-1)+....+ 2a-(n-1)+ 2a-(n-1)+2a-(n-1)
By principle of addition; a+a+a+a=4a we have that;
2Tn=n[2a-(n-1)]
©
©
©
©
now substitute 'n-1' for 'a' as in the case. of highest possible shakes. we have;
2Tn=n[2(n-1)-(n-1)]=n[n-1] (NB: 2a-a=a)
.
2Tn=n(n-1)
finally divide L.H.S and R.H.S of the equation by 2 to have Tn, we get;
2Tn/2=[n(n-1)]/2
therefore; Tn=n/2(n-1)

©®
CREDIT to my:
cracking BRAIN

Nimen:
wetin CJ find come here? come siddon for my side

efedua:
let n be the numbers of people placed on line to have a shakes with each others. obviously, the highest possible count of shakes,will be from the first person and must be given as 'n-1=a' and the lowest possible will be given as 'a-(n-2)=1'
however when we run a summation of shakes from nth term(being the first person shakes to others and left) to 1th term(being the last single person who will not shake any body and left) or a to 0 we will have an AP of a+a-1+a-2+....+a-(n-3)+a-(n-2)+a-(n-1)
.
(where a-(n-1)=0)
.
when we count the numbers of terms in the summation above we will get n numbers of terms.
prove:
when 3people have shakes, by the summation of a to 0, where n-1=a, we have the numbers of shakes as 2+1+0=3
now count the numbers of terms, we have 2,1 and 0 and they are three terms.

let write the summation from a to 0 as Tn, where Tn is the total numbers of shakes, we get;
Tn=a+a-1+a-2+....+a-(n-3)+a-(n-2)+a-(n-1)
.
Similarly, if we rewrite the summation from 0 to a as Tn*, we get;
Tn*=a-(n-1)+a-(n-2)+a-(n-3)+....+a-2+a-1+a. (where Tn*=Tn, Then Tn +Tn*=2Tn)
.
.
Now let add Tn to Tn* by adding there terms in the same position without affecting the n numbers of terms. that is, add Tn 'a' to Tn* 'a-(n-1)' and Tn 'a-(n-1)' to Tn* 'a' and do the same for all value that fall within them

for the formal we have;
a+a-(n-1)=2a+(n-1)
for the later we also have;
a-(n-1)+a=2a+(n-1)
NB:this constant value 2a+(n-1) will still occur for all added terms by position
prove:
from Tn 2nd term and Tn* 2nd term summation we get;
{a-1}+{a-(n-2)}= 2a-1-(n-2)=2a-1-n+2=2a-n+1=2a-(n-1)
.
.
.
.so therefore, the sum of Tn and Tn*, we give;
2Tn=2a-(n-1)+ 2a-(n-1)+ 2a-(n-1)+....+ 2a-(n-1)+ 2a-(n-1)+2a-(n-1)
By principle of addition; a+a+a+a=4a we have that;
2Tn=n[2a-(n-1)]
©
©
©
©
now substitute 'n-1' for 'a' as in the case. of highest possible shakes. we have;
2Tn=n[2(n-1)-(n-1)]=n[n-1] (NB: 2a-a=a)
.
2Tn=n(n-1)
finally divide L.H.S and R.H.S of the equation by 2 to have Tn, we get;
2Tn/2=[n(n-1)]/2
therefore; Tn=n/2(n-1)

©®
CREDIT to my:
cracking BRAIN

mesoade:
but guy how did
.
a+a-(n-1)=2a+(n-1) ,rather it ought be
a+a-(n-1)=2a-(n-1) . . If it should be what you wrote then there would be no needs for the brackets cos the expanded form will be 2a+n-1

Neldrizzy:
EmperorKing20 am so proud of you!!! Who ever believe you would go this far!!! MORALE : always believe in ur self and give a try no matter the circumstance(s).. Nevr Limit ur thinking........

Emperorking20:

Oga Nel! Thank u plenty 4 ur support.. It was invaluable
Really! Thanks..

nyeroluv:
Pls i'm a prospective aspiring uniben candidate...i can't find 2015/2016 thread anywhere...wil any1 link me pls if dere is any...so we cud knw ourselves beta...jamb is fast approaching!

Neldrizzy:
see as you don popular . Take more pics of ur Abs for the Final showdown, it will boost ur vote... i saw where a girl voted you today and she said she voted you cus of ur abs!

nyeroluv:
Pls i'm a prospective aspiring uniben candidate...i can't find 2015/2016 thread anywhere...wil any1 link me pls if dere is any...so we cud knw ourselves beta...jamb is fast approaching!

chue:
We use to av nite crawlers on dis thread where em?