how to find the diffrence in squares of two positive consecutive whole numbers without calculating them.
For example
(4*4) - (3*3)
2007-02-19 22:57:20 · 6 answers · asked by klark kent 1 in Science & Mathematics ➔ Mathematics
This is sum of the 2 numbers
basis
(n+1)^2 - n^ 2= n^2 +2n +1 - n^2 = 2n+1 = (n+1) + n
2007-02-19 23:00:48 · answer #1 · answered by Mein Hoon Na 7 · 1⤊ 0⤋
(n+1)^2-n^2 = (n+1+n)(n+1-n) = n +n+1 which is the sum of the numbers
2007-02-20 07:07:01 · answer #2 · answered by santmann2002 7 · 1⤊ 0⤋
Hey Clark,
Its simple,
U jus have to add the two numbers and that is the difference between the squares of two consecutive whole numbers. Check it out for urself.
2007-02-20 07:03:08 · answer #3 · answered by death approaching 1 · 1⤊ 1⤋
let the number be 'n'
Required:n^2-(n-1)^2
Expand using formula (a^2)-(b^2)=(a+b)(a-b)
Ans: (n+(n-1)) (n-(n-1))=2n-1
2007-02-21 04:05:25 · answer #4 · answered by Aneeqa 4 · 0⤊ 0⤋
just add those two number...
(4*4)-(3*3)=7...4+3
(10*10)-(11*11)=21...10+11
2007-02-20 09:56:45 · answer #5 · answered by jrpatel 3 · 0⤊ 0⤋
a^2-b^2=(a+b)(a-b)
Use the above formula.
2007-02-20 07:11:05 · answer #6 · answered by Bishu 3 · 1⤊ 0⤋