Summing A Cool Infinite Series

[@Marcus-y1m]

4:17 mind blowing 🤯 . That was a genius observation. Thanks sir!

[@ZachariusDackley]

I have a question. What are the key concepts i have to study to learn to solve this ? I am willing to review them first so that i could master this problem from the video. I am now starting to study math through KZbin videos. A reply may help me a lot. This is the 3rd video i am watching of your channel. I appreciate your work Sir !

[@nothinghere9576]

This solution uses differentiation, if you want to understand it, you should know about derivatives and antiderivatives, you study those in calculus. You also need formula for the sum of geometric series, you can review what geometric series is in general.

[@sonicbreaker00]

What grade/class are you in ?

[@MrGeorge1896]

We can find the term for the sum of (n + 1)² * x^n even w/o differentiation just with the basic formula 1 + x² + x³ + ... = 1 / (1 - x). So 1 + 4x + 9x² + 16x³ + ... = (1 + 3x + 5x² + 7x³ + ...) * (1 + x + x² + x³ + ...) = (1 + 3x + 5x² + 7x³ + ...) / (1 - x) = (1 + 2x + 2x² + 2x³ + ...) * (1 + x + x² + x³ + ...) / (1 - x) = (1 + 2x + 2x² + 2x³ + ...) / (1 - x)² = ( -1 + 2 * (1 + x + x² + x³ + ...) ) / (1 - x)² = (-1 + 2 / (1 - x) ) / (1 - x)² = (x - 1 + 2) / (1 - x)³ = (x + 1) / (1 - x)³

[@doctorb9264]

Excellent work on these sum problems.

[@guyhoghton399]

Sometimes it helps to multiply and then take the difference with series like these. Let *_S = 1²/7⁰ + 2²/7¹ + 3²/7² + ... + (n + 1)²/7ⁿ +_* ... ∴ _7S = 7 + 2²/7⁰ + 3²/7¹ + ... + (n + 2)²/7ⁿ +_ ... Taking the difference: _6S = 7 + (2² - 1²)/7⁰ + (3² - 2²)/7¹ + ... + [(n + 2)² - (n + 1)²]/7ⁿ +_ ... ⇒ _6S = 7 + Σ₀∞{(2n + 3)/7ⁿ}_ ⇒ *_6S = 7 + 3Σ₀∞{1/7ⁿ} + 2Σ₀∞{n/7ⁿ}_* ... ① This has decomposed _S_ into a constant, a G.P. and a series that can be derived from the derivative of a G.P. Let *_f(x) = Σ₀∞{(x/7)ⁿ} = 1/(1 - x/7) = 7/(7 - x)_* ∴ _f(1) = Σ₀∞{1/7ⁿ}_ _f'(x) = 7/(7 - x)² = Σ₀∞{nxⁿ⁻¹/7ⁿ}_ ∴ _f'(1) = Σ₀∞{n/7ⁿ}_ From ①: _6S = 7 + 3f(1) + 2f'(1)_ ⇒ *_S = ⅙{7 + (3)(7/6) + (2)(7/36)} = 49/27_*

[@paulortega5317]

In general n²(n+1)/(n-1)³, in this case n=7

[@ciaucia156]

S - rS = S₁ = 1 + 3r + 5r² + 7r³ +… S₁- rS₁= S₂ = 1 + 2r + 2r² + 2r³ +… = 1 + 2r(1 + r + r² +…) = 1 + 2r/(1-r) S₁(1-r) = S(1-r)² = (1+r)/(1-r) S = (1+r)/(1-r)³

[@msmbpc24]

Nice trick.

[@SyberMath]

Thanks

[@russellsharpe288]

s = 1/1 + 4/7 + 9/49 + 16/343 +.... s/7 = 1/7 + 4/49 + 9/343 + ... subtract two previous lines to find 6s/7 = 1/1 + 3/7 + 5/49 + 7/343 +... 6s/49 = 1/7 + 3/49 + 5/343 + ... subtract two previous lines to find 36s/49 = 1/1 + 2/7 + 2/49 + 2/343 + ... now subtract 1 and use standard geometric sum to find (36s - 49)/49 = 2/7 (1+ 1/7 + 1/49 +...) = 2/7 x 7/6 = 1/3 36s - 49 = 49/3 36s = 49 x 4/3 s = 49/27

[@RashmiRay-c1y]

The sum is d/dr r d/dr [1/(1-r) ] = (1+r)/(1-r)^3 = 49/27.

[@Evil_Jyan]

16/91

[@doctorb9264]

The first term is already 1.

[@Evil_Jyan]

@doctorb9264 I meant 343

[@chathurangasameera6722]

I have an nother idea we get the fair formula of sequence ur then ur=((1+r)^2)/7^r i allready foget the first term is 1 okay then my ur is like without 1 okay my target is sigma r goes to 1 to infinity ur + 1. This is the sum of the sequence.okay then after i suppose ur = f(r)-f(r+1) Then f(r)=(Ar^2+Br+C)/7^r And finally you can substitute f(r) and f(r+1) and you can get the values of A,B,C . A=7/6 B=49/18 C=49/27 Then after we can get sum of ur Okay r=1 u1 = f1-f2 r=2 u2 = f2 - f3 furtherly r=n-1 un-1 =f(n-1)-f(n) r=n un = f(n) - f (n+1) Okay adding all this we can get sigma r running to 1 to n ur = f1-f(n+1) Okay finally we get the limits of n goes to infinity f1-f(n+1) value Easily to calculate its value is 22/27 Finally sum is given 1+22/27=49/27.

[@giuseppemalaguti435]

S=Σk^2/7^(k-1)..k=1,2,3...=7Σk^2(1/7)^k=7(1/7)(1+1/7)/(1-1/7)^3=7(8/49)/(216/343)=49/27

[@scottleung9587]

Nice!

[@Don-Ensley]

problem Can you sum 1 + 4 / 7 + 9 / 49 + ... Let the nth term be Sₙ. Examination of the pattern reveals Sₙ. Sₙ = ( n + 1 )² / ( 7 ⁿ ) , with n starting at 0. The ratio test for convergence is used. lim (Sₙ₊₁/Sₙ)=(1/7)[(n+ 2)²]/[(n+1)²] n→∞ = 1/7 < 1 This series converges absolutely. Let the common ratio be r. r = 1 / 7 Sₙ = ( n + 1 )² r ⁿ Use sigma notation to denote the sum S. ͚ S = Σ ( n + 1 )² r ⁿ ⁿ⁼⁰ This is best tackled by an expansion of Sₙ. Sₙ = ( n + 1 )² r ⁿ Sₙ = (n² + 2 n + 1)r ⁿ Split the series up into three pieces. ͚ ͚ ͚ S = Σ n² r ⁿ + 2 Σ n r ⁿ + Σ r ⁿ ⁿ⁼⁰ ⁿ⁼⁰ ⁿ⁼⁰ Let P₁, P₂, P₃ represent the n², n, and unity series terms. Use the known sum of a geometric series formula as the given. ͚ P₃ = Σ r ⁿ = 1 / ( 1 - r ) ⁿ⁼⁰ = 7 / 6 Take the derivatives on each side with respect to r. ͚ Σ n r ⁿ / r = 1 / ( 1 - r )² ⁿ⁼⁰ Multiply by r to find P₂. ͚ P₂ = Σ n r ⁿ = r / ( 1 - r )² ⁿ⁼⁰ = 7 / 36 ͚ Σ n r ⁿ = r / ( 1 - r )² ⁿ⁼⁰ Take the derivatives again on each side with respect to r. ͚ Σ n² r ⁿ / r = d[ r / (1 - r)² ]/dr ⁿ⁼⁰ = (1 + r) / [(1 - r)³ ] Multiply by r to find P₁. ͚ P₁ = Σ n² r ⁿ = ( r + r² ) / [ ( 1 - r )³ ] ⁿ⁼⁰ = 56 / 216 S = P₁ + 2 P₂ + P₃ = (56 / 216) + (84/ 216) + (252 / 216) = 392 / 216 = 49 / 27 answer 49 / 27

[@yurenchu]

_Answer_ : 49/27 _Calculation_ : Sum S = 1 + 4/7 + 9/49 + ... + k²/7^(k-1) + ... Let f(x) = 1 + 4x + 9x² + ... Then S = f(1/7) and ∫ f(t) dt , from t=0 to t=x = x + 2x² + 3x³ + ... = x*(1 + 2x + 3x² + ... ) = x*g(x) where g(x) = (1 + 2x + 3x² + ... ) = = d/dx (x + x² + x³ + ...) = d/dx (x/(1-x)) = ((1-x) + x)/(1-x)² = 1/(1-x)² ∫ f(t) dt , from t=0 to t=x = x*g(x) = x*(1/(1-x)²) = x/(1-x)² f(x) = d/dx ( x/(1-x)² ) = [(1-x)² + x*2(1-x)]/(1-x)⁴ = = [(1-x) + 2x]/(1-x)³ = [1 + x]/(1-x)³ S = f(1/7) = = [1 + 1/7]/(1 - 1/7)³ = [8/7]/(6/7)³ = [8/7]/(6³/7³) = [8*7²]/(6³) = [7²]/(3³) = 49/27

[@Marcus-y1m]

Good Job

[@ZachariusDackley]

With that , it becomes more difficult for me . But thanks