33So 4 methods 4 this problem...
(3 pink and 1 white)
:)
derive the formulae for
12+22+32+......+n2
without using mathematical induction
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22 Answers
62hmm.. this is an interesting proof.. good that you posted it after finding :)
Well if u got the answer and u r convinced enuf, can there be errors ;)
33Yippie that Σr2 is also done:
n*n+1C2-n+1C3
So the method is correct...
Yippies so it should be pink....
[1] [1] [1] [1]
62@Akshay: you are right.. (almost absolutely)
But if u see some of the methods posted... they are very interesting... and they give u good insight into such problems.... It is not about learning stuff.. there are some tricks here that can be useful..
33But its fun to derive these......
(Edited something....above)
1NCERT 11th class book
came in our school paper in 11th,I remember
But we have to remember it,we can't take this as a question,this is a derivation
33Yippie..
Solved and got the answer for Σr and answer coming is correct
62i am struggling hard to understand ur method..
is it due to some stuff that i dont know.. or am i sleeping :D
33~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
One other method can be...
Write the summation as ΣrrC1
Coefficient of x1 in 0(1+x)0+1(1+x)1+2(1+x)2+......
(as it is 0*0+1*1+2*2+3*3+... which is the req sum
and it is the coefficient of x1 in the expansion of Σr(1+x)r
Sum can be found as it is AGP or something like that..
i.e. it is the coefficient of x1 in the expansion of
(1+x)/x[n(1+x)n-{(1+x)n-1}/x]
or the coefficient of x3 in the expansion of (1+x){nx(1+x)n-(1+x)n}
...
Similarly for Σr
it can be written as ΣrrC0
Which is the coefficient of x0 in the expansion of
(1+x)/x[n(1+x)n-{(1+x)n-1}/x]
Or the coefficient of x2 in the expansion of
(1+x){nx(1+x)n-(1+x)n}
..........
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Italics are edits..
62Note that
24r² + 2 ≡ (2r+1)³ - (2r-1)³
24(1²) + 2 = 3³ - 1³
24(2²) + 2 = 5³ - 3³
24(3²) + 2 = 7³ - 5³
24[(n-1)²] + 2 = (2n-1)³ - (2n-3)³
24[(n)²] + 2 = (2n+1)³ - (2n-1)³
Adding all of these gives:
24[1²+2²+3²+...+(n-1)²+n²] + 2n = (2n+1)³ - 1³
That is: 24 [sum of all squares up to n²] + 2n = 8n³ + 12n² + 6n
24[sum of all squares up to n²] = 8n³ + 12n² + 4n
= 4n(2n² + 3n + 1)
[sum of all squares up to n²] = (n/6)(2n² + 3n + 1)
= (n/6)(2n+1)(n+1)
1i got sumwhere using my technique
please correct m if i go wrong
S = 12+22+.....+n2
we know every sqaure can be written as the sum of first n natural odd numbers
1=1
4=1+3
9=1+3+5
so on
S = 1+(1+3)+(1+3+5)+.....
S = n*1 + (n-1)*3 + (n-2)*5 + ....... +(n-(n-1))(2n-1)
* 1 appears n times
3 appears n-1 times and so on
S = n*1 + (n-1)*3 + (n-2)*5 + ....... +(1)(2n-1)
now plz try and understand SUBTRACT n from first term (n-1) from second term (n-2) frm third term and so on and 1 from the last term
IT IS LIKE REDUCING THE COEFFICIENT OF EACH TERM BY 1.
S = 0+2(n-1)+4(n-2)+6(n-3)+.......+1(2n-2) +
(n+n-1+n-3+n-4...)
the (n+n+1+n+2+...) is like sum of first n natural numbers frm the back side and hence can be replaced by n(n+1)/2
S = 2(n-1)+4(n-2)+6(n-3)+....+(2n-1) + n(n+1)/2
take 2 common now on rhs forget the n(n+1)/2 for time being
2(n-1+2n-4+3n-9+4n-16+.....)
=2(n+2n+3n+4n+..... - S)
S is what we have to find
=2 ( n2(n+1)/2 -S)
opening the bracket
2n2(n+1)/2 - 2S + n(n+1)/2
S = 2n2(n+1)/2 - 2S + n(n+1)/2
3S = 2n2(n+1)/2+ n(n+1)/2
3s=((n+1)(2n2+n))/2
on simplifying or factorising
3S=n(n+1)(2n+1)/2
therefore
S=n(n+1)(2n+1)/6
plz feel free to point any mistakes if found i would be greatful
341For greatvishal:
You would have solved this problem:
1/1.2 + 1/2.3 + 1/3.4+......1/k(k+1)+...+1/(n-1)n
by writing the general term as Tk = 1/k(k+1) = 1/k - 1/k+1
So, we get the sum to be equivalent to
(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-2 - 1/n-1) +(1/n-1 - 1/n)
You can easily see how the intermediate terms get knocked off in successive brackets leaving only
1-1/n = n-1/n
This is known as telescopic sum
In general we write it as Σf(r) - f(r-1)
Its like the barrels of an amateur telescope, they go into each other successively till it becomes small enough to be carried in the pockets of the great coats that used to be the fashion in Victorian times!
Other instances would be
S = sin(a) + sin(a+h)+....+sin(a+(n-1)h)
S* sin h/2 is a telescopic sum
62vishal dont worry aobut that word.. i think the answer is pretty clear even when u eliminate that word :)
341Another way would be:
Consider Σk2+k
This can be done in two ways
1. Σk2+Σk
2.Σk(k+1)
The second one can be written as a telescopic sum as
ΣTr
where Tr = 1/3 [r(r+1)(r+2) - (r-1)r(r+1)]
so the sum is n(n+1)(n+2)/3
Hence Σk2+Σk = n(n+1)(n+2)/3
So that Σk2 = n(n+1)(n+2)/3 - Σk
= n(n+1)(n+2)/3 - n(n+1)/2
= n(n+1)(2n+1)/6
62well indraneel i love the way you have thought...
It shows a lot ...
But for the last 10 mins i have tried to think on this one and some how come to realise or feel! (I hope someone proves me wrong and finds a way to solve this the way you have put it!)
But i feel that this cannot be solved by this method because the summation that you put this way will also have summation of squares....
Atleast I could not get across ... I hope sincerely someone gets this one so that we can see a brilliant thought put to a brilliant conclusion
1thanks.
but can you tell me whether we can proceed using this or no
1=1
4=1+3
9=1+3+5
16=1+3+5+7
n2=1+3+5+7+.....+2n-1
1oh my god.............excellent method nishant bhaiyya, i wud never even thought of such a method even in my wildest dreams............
62The problem is that this is something you have to know i dont think if it can come naturally to most people...
It can if you see carefully that we need to take it as the difference of the fields.....
Having said that it is not very easy to do!