how do we find the sum of the terms after the n^(th) term of a GP


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 163675 by alcohol last updated on 09/Jan/22

$h o w d o w e f i n d t h e s u m o f t h e t e r m s a f t e r$ $t h e n^{t h} t e r m o f a G P$
Commented by mr W last updated on 09/Jan/22

$m a k e t h e {(n + 1)}^{t h} t e r m t o t h e f i r s t$ $t e r m a n d a p p l y t h e f o r m u l a f o r s u m$ $f r o m f i r s t t e r m t o \infty .$ $a_{1} + a_{1} q + a_{1} q^{2} + . . . + a_{1} q^{n - 1} + a_{1} q^{n} + a_{1} q^{n + 1} + . . . = \frac{a_{1}}{1 - q}$ $a_{1} q^{n} + a_{1} q^{n + 1} + . . . = \frac{a_{1} q^{n}}{1 - q}$ $o r$ $s u m f r o m f i r s t t o n^{t h} t e r m : \frac{a_{1} (1 - q^{n})}{1 - q}$ $s u m f r o m f i r s t t e r m t o \infty : \frac{a_{1}}{1 - q}$ $s u m f r o m {(n + 1)}^{t b} t e r m t o \infty : \frac{a_{1}}{1 - q} - \frac{a_{1} (1 - q^{n})}{1 - q} = \frac{a_{1} q^{n}}{1 - q}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum