sum the series to n terms and infinite terms of progression 1+ 4/5+7/5²+10/5³........... is
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S∞ = 1 + 4/5 + 7/5² + 10/5³ + .............................
S∞/5 = 1/5 + 4/5² + 7/5³ + .......................
- - - -
S∞(1-1/5) = 1 + 3/5 + 3/5² + 3/5³ + .............
4S∞/5 = 1 + 3/5(1 + 1/5 + 1/5² + .........................................)
4S∞/5 = 1 + 3/5(1/(1-1/5))
4S∞/5 = 1 + (3/5)(5/4)
4S∞/5 = 7/4
S∞ = 35/16
4Sn/5 = 1 + (3/5)[1{(1-1/5^n)/(1-1/5)}]
4Sn/5 = 1 + (3/5){5/4(1-1/5^n)]
4Sn/5 = 1 + (3/4)(1-1/5^n)
Sn = 5/4 + (15/16)(1-1/5^n)
or
Sn = 5/4 + (3/16)(5^n-1)/5^n-1
i hope you can understand .
S∞/5 = 1/5 + 4/5² + 7/5³ + .......................
- - - -
S∞(1-1/5) = 1 + 3/5 + 3/5² + 3/5³ + .............
4S∞/5 = 1 + 3/5(1 + 1/5 + 1/5² + .........................................)
4S∞/5 = 1 + 3/5(1/(1-1/5))
4S∞/5 = 1 + (3/5)(5/4)
4S∞/5 = 7/4
S∞ = 35/16
4Sn/5 = 1 + (3/5)[1{(1-1/5^n)/(1-1/5)}]
4Sn/5 = 1 + (3/5){5/4(1-1/5^n)]
4Sn/5 = 1 + (3/4)(1-1/5^n)
Sn = 5/4 + (15/16)(1-1/5^n)
or
Sn = 5/4 + (3/16)(5^n-1)/5^n-1
i hope you can understand .
knight:
this is sum of infinite terms..... bt what is sum of n terms
sorry it is just my mistake .
sry its nt matching u hv made a mistake
u hv taken n terms in Sum formula bt its only n-1 terms....... btw i got the ans
thankew fr efforts :)
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so this is the correct answer
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