geometric sum formula Sum of Geometric Series Formula duplicate Ask Question Asked 6 years 3 months ago Modified
We know that the formula for computing a geometric series is sum i 1 infty a 0r i 1 frac a 0 1 r Out of curiosity I would like ask Is there any ways the formula can be derived o begingroup I had always learned the formula as the first term over 1 r In that case you can start with any index I suppose your answer illustrates why this is true endgroup
geometric sum formula
geometric sum formula
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Derivation Of The Formula For The Sum Of A Geometric Series YouTube
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How To Find The Sum To Infinity Of A Geometric Series Mathsathome
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1 Answer Sorted by 3 Notice the following steps Step 1 setting n 1 n 1 we get 1 r 1 r2 1 r 1 r 1 r 1 r 1 r 2 1 r 1 r 1 r Step 2 assuming it holds for n k n k then 1 r r2 rk 1 rk 1 1 r 1 r r 2 r k 1 r k 1 1 r Step 3 substituting n k 1 n k 1 Nov 16 2015 at 14 56 the sum of geometric series is given by a rn 1 r 1 a r n 1 r 1 here r x a x r x a x Archis Welankar Nov 16 2015 at 14 57 I am not surprised that you got strange results for x 2 x 2 since you re in that case trying to calculate 2 4 8 16 2 4 8 16 which
Modified 8 years 6 months ago Viewed 412 times 2 The formula to find sum of n terms of a geometric progression is a rn 1 r 1 a r n 1 r 1 if r 1 r 1 a 1 rn 1 r a 1 r n 1 r if r Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow the largest most trusted online community for developers to learn share their knowledge and build their careers
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The case an 2n a n 2 n is the most intuitive to me because 2n 2n 2 1 2n 1 2n 2 n 2 n 2 1 2 n 1 2 n Then noticing the telescoping sum the formula easily follows If we try to do the same kind off thing with x x instead of 2 2 we get xn xn x 1 x 1 xn 1 xn x 1 x n x n x 1 x 1 x n 1 Proving the geometric sum formula by induction Ask Question Asked 10 years 9 months ago
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geometric sum formula - 1 Answer Sorted by 3 Notice the following steps Step 1 setting n 1 n 1 we get 1 r 1 r2 1 r 1 r 1 r 1 r 1 r 2 1 r 1 r 1 r Step 2 assuming it holds for n k n k then 1 r r2 rk 1 rk 1 1 r 1 r r 2 r k 1 r k 1 1 r Step 3 substituting n k 1 n k 1