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Question Number 121949 by liberty last updated on 12/Nov/20

Evaluate the sum (2/(3+1)) + (2^2 /(3^2 +1)) + (2^3 /(3^4 +1))+ ...+ (2^(n+1) /(3^2^n +1)) .

Evaluatethesum23+1+2232+1+2334+1+...+2n+132n+1.

Answered by bobhans last updated on 12/Nov/20

consider (1/(a+1))= ((a−1)/(a^2 +1))=(a/(a^2 −1))−(1/(a^2 −1)) =−(1/(a^2 −1)) + (1/2)((1/(a+1))+(1/(a−1))) which implies (1/2).(1/(a+1))=(1/(2(a−1)))−(1/(a^2 −1)) substitute identity with a=3^2^k we obtain (1/(2(3^2^k +1))) = (1/(2(3^2^k −1)))−(1/(3^2^(k+1) −1)) multiplying this by 2^(k+2) , we get the relation (2^(k+1) /(3^2^k +1)) = (2^(k+1) /(3^2^k −1))−(2^(k+2) /(3^2^(k+1) −1)) Thus (2/(3+1))+(2^2 /(3^2 +1))+(2^3 /(3^4 +1))+...+(2^(n+1) /(3^2^n +1)) = Σ_(k= 0) ^n ((2^(k+1) /(3^2^k −1)) −(2^(k+2) /(3^2^(k+1) −1)) ) = 1−(2^(n+2) /(3^2^(n+1) −1)).

consider1a+1=a−1a2+1=aa2−1−1a2−1=−1a2−1+12(1a+1+1a−1)whichimplies12.1a+1=12(a−1)−1a2−1substituteidentitywitha=32kweobtain12(32k+1)=12(32k−1)−132k+1−1multiplyingthisby2k+2,wegettherelation2k+132k+1=2k+132k−1−2k+232k+1−1Thus23+1+2232+1+2334+1+...+2n+132n+1=∑nk=0(2k+132k−1−2k+232k+1−1)=1−2n+232n+1−1.

Commented by liberty last updated on 12/Nov/20

good....

good....

Commented by sewak last updated on 17/Nov/20

Quite good

Quitegood

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