If: 7 = 1 1 X + 1 2 1 X + 1 3 3 1 X + 1 4 6 4 1 X + ⋯ t o ∞ Find the value of X
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7 = 1 1 1 X + 1 1 2 X + 1 1 3 X + 1 1 4 X + ⋯ → n = 1 , 7 = 1 1 n X + 1 1 n + 1 X + 1 1 n + 2 X + ⋯ . Then lets get a approximation of 7 = 1 1 X . Maybe it is 7 7 = X . Next, 7 = 1 1 2 X → 8 4 7 = X . And so on.... Then, let the final number A . Next, A = 1 1 b X , b ≥ 1 . Then 7 × 1 1 b − 1 = ( b − 1 ) X + ( b − 2 ) X + ( b − 3 ) X + ⋯ . Although we do not know the value of b , but lets keep calculating. In 7 = 1 1 X + 1 2 1 X + 1 3 3 1 X + 1 4 6 4 1 X , we get 1 0 2 4 8 7 = 1 3 3 1 X + 1 2 1 X + 1 1 X + X . Then 1 4 6 4 X = 1 0 2 4 8 7 . Lets finish the calculating with X = 1 4 6 4 1 0 2 4 8 7 . Lets round it into the integer form. Its approximation is 7 0 , so the answer is 7 0 .
n = 0 ∑ ∞ a r n = 1 − r a
7 = 1 1 X + 1 2 1 X = 1 3 3 1 X + 1 4 6 4 1 X + ⋯
a = 1 1 X , r = 1 1 1
n = 0 ∑ ∞ x A ⋅ ( x B ) n = x − B A
A = X , B = 1 , x = 1 1
n = 0 ∑ ∞ 1 1 X ⋅ ( 1 1 1 ) n = 1 1 − 1 X = 1 0 X
7 = 1 0 X , X = 7 0
X = 7 0
7 = X ( 1 1 1 + 1 2 1 1 + 1 3 3 1 1 + 1 4 6 4 1 1 + ⋯ ) s = ( 1 1 1 + 1 2 1 1 + 1 3 3 1 1 + 1 4 6 4 1 1 + ⋯ ) 1 1 s = 1 + 1 1 1 + 1 2 1 1 + 1 3 3 1 1 + 1 4 6 4 1 1 + ⋯ 1 1 s = 1 + s ⟶ s = 1 0 1 7 = X s ⟶ 7 = 1 0 X ⟶ X = 7 0 note that n = 1 ∑ ∞ S n 1 = S − 1 1
Quite easy. Just use the formula for infinite sum of a GP.
So what is it? I let you explain that.
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Let 7 = S .Then: S = 1 1 X + 1 2 1 X + 1 3 3 1 X + 1 4 6 4 1 X + ⋯ 1 1 S = 1 2 1 X + 1 3 3 1 X + 1 4 6 4 1 X + ⋯ S − 1 1 S = ( 1 1 X + 1 2 1 X + ⋯ ) − ( 1 2 1 X + 1 3 3 1 X + ⋯ ) 1 1 1 0 S = 1 1 X 1 0 S = X Plugging in S = 7 ,We get: 1 0 ( 7 ) = X = 7 0