Perfect square problems |
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Number pattern 1
´ 2 ´ 3 ´ 4 + 1 =
25 = 52 2
´ 3 ´ 4 ´ 5 + 1 =
121 = 112 3
´ 4 ´ 5 ´ 6 + 1 =
361 = 192 5
´ 6 ´ 7 ´ 8 + 1 =
841 =
292 : : : : : |
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Algebra The
pattern seems to work onwards. Can we prove this? Simple!
Use algebra! n(n
+ 1)(n + 2)(n + 3) + 1 =
[n(n + 3)] [(n + 1)(n + 2)] + 1 =
[n2 + 3n] [n2 + 3n + 2] + 1 =
[(n2 + 3n + 1) – 1] [(n2 + 3n + 1) + 1] + 1 =
(n2 + 3n + 1)2 – 12 + 1 [since ( x – y )(x + y) = x2 – y2] =
(n2 + 3n + 1)2,
which
is a complete square. \ n(n + 1)(n + 2)(n + 3) +
1 = (n2 + 3n + 1)2 If
you put in n = 1, 2, 3, 4, …. ,
you can get the number pattern on the top. |
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49 = 72 4489 = 672 444889 = 6672 44448889 = 66672 : : : : |
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Algebra Let m = 11….1 (n times) \ 10n = 9 × 11….1 + 1 = 9m + 1 (n
times) 44….……..488……..…..89
= 44….…4 ×10n +
88…..…8 + 1 (n
times) (n-1
times) (n
times) (n
times) = 4m (9m + 1) +
8m + 1 =
36 m2 + 4m
+ 8 m + 1 =
36 m2 +
12m + 1 =
(6m + 1)2 = (66………6 + 1)2 (n
times) =
(66…..…..67)2 (n-1
times) |
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Number pattern
11
–
2 = 9 = 32 1111
–
22 =
1089 =
332 111111
–
222 = 110889 =
3332 11111111
– 2222 = 11108889 = 33332 : : : : : |
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Algebra Let m
= 11….11 (n times) Then 22….22
(n times) =
2m 11….11 (2n
times) = 11….11 (n
times) ´ 100….00 (n
zeros) + 11….11 (n times) =
100….00 (n zeros) ´ m + m \ 11….11 (2n times) – 22….22 (n times) =
[100….00 (n zeros) ´ m + m] – 2m = 100….00 (n
zeros) ´ m – m = 99….99 (n
times) ´ m = 9m ´ m = (3m)2 = (33….33)2 (n threes) |