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1, the sum upto n terms is a perfect square.
Comments (13)
1st question we draw a circle of radius a/4 ,a being the length of the side so the points within the circle are closer to it's centre then the sides from that you can easily derive probability,the centre of circle is obviously the centre of the square.
wrong solution i shouldn't be rated
This can be made a little more rigorous as we still dont know if there are any other families of APs that do not have this property.
For this you have to look to the theory of finite differences.
What kind of sequences generate squares and are summations of an Arithmetic Progression, i.e. the difference of consecutive terms in the given sequence is an AP. From finite differences, this is a quadratic sequence. The fact that every sequence is a square means, the sequence in general is of the form 
I will write it this way
f: t1 t2 t3 t4 ......
AP: t1 (t2-t1) (t3-t2) (t4-t3).......
From this we see that the first term of both sequences has to be (a+b)2 = a2+2ab+b2.
Now, the general term of the A.P. is ![f(n)-f(n-1) = (an+b)^2 - [a(n-1)+b]^2 =(2ab-a^2)+2a^2n](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/d/b/9/db97c8fce5bea8f2aa614effa4888d0ce63ae4cc.gif)
So the first term of the A.P. is 
So, we must have 
from which we obtain b = 0.
That means the general term of the A.P. is 
. which is the sequence 
obtained by Ankit
a general solution:
let first term and the common difference be a,d respectively
so we have n[2a+(n-1)d] 2 = k2 ~~~~~~~~~~~~~~~(1)
for integral RHS we must have an even d so let d=2p
LHS = n[a+np-p] = pn2 + n(a-p) + 0
since LHS is a perfect square, the polynomial pn2 + n(a-p) + 0 must have equal roots
so discriminant = (a-p)2 = 0 or a = p
so d=2a
substituting for d in (1)
we have an2 = k2
or 'a' must be a perfect square. so in general the rule is
d=2a and a=perfect square
@mirka actually my above solution is wrong. i messed up with odd and even cases 
. here's another
let first term and the common difference be a,d respectively
so we have n[2a+(n-1)d] / 2 = k2 ~~~~~~~~~~~~~~~~~~~~(1)
or n[2a+(n-1)d] = [root(2) k]2
lhs= dn2 + n(2a-d)+0
as RHS is a perfect square, we must have equal roots of the polynomial dn2 + n(2a-d)+0
or (2a-d)2=0 or 2a=d
putting n=1 in (1) we get a=k2
or the AP is k2+3k2+5k2+...
as obtained by bhatt sir and ankit
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Ans 2
Sum of n consecutive odd numbers starting from '1' is always a perfect square.
1 + 3 + 5...... + 2n+1 = n^2
We can check this using formula for sum of AP to n terms.
Alternatively
22-12 = (3)(1)
32-22= (5)(1)
42-32= (7)(1)
52-42=(9)(1)
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(n)2 - (n-1)2 = 2n+1
Adding all equations we get LHS = n2 and RHS as 1+3+5+....2n+1
I call this the bomb method as everything gets cancelled