Find the sum S of infinite sequence of real numbers.
The sequence of numbers converges.
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The answer is 5.
The solution is in the enclosed picture.
The solution is in the enclosed picture.
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kvnmurty:
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a_0 = 1.
a_{n+1} = sqrt[ 4 + 3 × a_n + (a_n)^2 ] - 2, for n >=0.
From the given equation:
(a_{n+1} +2)^2 = 4 + 3 × (a_n )+ (a_n)^2.
Simplify.
Write this equation for n = 0 to N.
Sum up all equations and cancel equal terms from LHS and RHS ..
We get that a_N =0 as N -> infty, because (a_N+1)/(a_N) < 1... And a_0=1.
We get Sigma {n=1 to infty } a_n = 4...
Add a_0. We Get 5.
a_{n+1} = sqrt[ 4 + 3 × a_n + (a_n)^2 ] - 2, for n >=0.
From the given equation:
(a_{n+1} +2)^2 = 4 + 3 × (a_n )+ (a_n)^2.
Simplify.
Write this equation for n = 0 to N.
Sum up all equations and cancel equal terms from LHS and RHS ..
We get that a_N =0 as N -> infty, because (a_N+1)/(a_N) < 1... And a_0=1.
We get Sigma {n=1 to infty } a_n = 4...
Add a_0. We Get 5.
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