What is the gradient of y = 1 + x + x 2 + x 3 . . . x n … when x = 2 1 .
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Same solution :D
To find the gradient, we have to differentiate the equation:
y = 1 + x + x 2 + x 3 … x n …
d x d y = 0 + 1 + 2 x + 3 x 2 … n x n − 1 …
Then, we substitute x = 2 1 and solve the resulting geometric series:
d x d y = 1 + 2 ( 2 1 ) + 3 ( 2 1 ) 2 … n ( 2 1 ) n − 1 …
2 × d x d y = 2 + 2 + 3 ( 2 1 ) + 4 ( 2 1 ) 2 … n ( 2 1 ) n − 2 …
We subtract the first equation from the second equation:
d x d y = 3 + 2 1 + ( 2 1 ) 2 + ( 2 1 ) 3 … ( 2 1 ) n ⋯ = 3 + 1 = 4
by differentiating we get 1+2x+3x^2+................................+nx^(n-1) by binomial theorem it is equal to (1-x)^(-2) putting x =1/2 we get 4
y = "unlimited pattern" 1/1-x = 1/1-1/2 = 1/(1/2) = 2. Because gradient (m) = y/x. So, the gradient is 2/(1/2)=4. Answer : 4
y = 1 + x + x 2 + x 3 + ⋯ + x n + ⋯ ∞ = 1 − x 1 ⇒ y ′ = ( 1 − x ) 2 1 ⇒ y ′ ( 2 1 ) = 4 .
However, y as shown in the question is an infinite power series. I am unsure if it can be treated as a function. At most we can say, y ′ ( 2 1 ) → 4 as n → ∞ .
dy=1+2x+3x^2+4x^3+................+...........to infinity. After Summing up the series we will get 1/(1-x)^2.Put x=1/2 and get the ans.
The sequence can be write down as y = 1 / ( 1 − x ) , and the gradient can be found by finding the derivative of this function.
The derivative: y ′ = 1 / ( 1 − x ) 2
Put x = 1 / 2 to the derivative to get a gradient: y ′ = 4
maka yet hota pan muddam sanguk nay
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Note the identity 1 + x + x 2 + … = 1 − x 1 , which follows from the formula for a geometric series. (Since x = 1 / 2 , the sum converges.) Then, we have, by the Quotient Rule, d x d ( 1 − x 1 ) = ( 1 − x ) 2 1 ( 1 ) − ( 1 − x ) ( 0 ) = ( 1 − x ) 2 1 so when x = 1 / 2 , the gradient of y is ( 1 − 1 / 2 ) 2 1 = 4 .