(In reply to
Puzzle Solution: Method 2 by K Sengupta)
Let the nth term of the given sequence be T(n) and the sum to the mth term of the given sequence be denoted by S(n).
Now, we have:
a= first term of the given sequence =1
d= common difference =d
Therefore,:
T(n) = a+(n-1)d=1+(n-1)2=2n-1
Now, we observe that:
S(1)=T(1)=1=1^2
S(2)=T(2)=1+3=4=2^2
S(3)=T(3)= 1+3+5=9=3^2
This leads one to conjecture that:
S(n)=n^2
Let the above relationship be satisfied for n= p
We will prove that the said relationship is also satisfied for n=p+1
Since the relationship is satisfied for n=p, we must have:
S(p)=p^2
Then, we must have:
S(p+1)
= S(p)+T(p+1)
= p^2 +2(p+1)-1
=p^2+2p+1
=(p+1)^2
Hence, the above relationship is also satisfied for n=p+1
Accordingly, we have proven that S(n)=n^2 and consequently the sum of consecutive odd numbers always add up to a perfect square.
Edited on May 11, 2022, 9:50 pm
Puzzle Solution: Method 3
