Last updated at May 29, 2018 by Teachoo
Transcript
Ex9.3, 11 Introduction (2 + 3k) At k = 1, 2 + 31 At k = 2, 2 + 32 .. โฆ โฆ. At k = 11, 2 + 311 Ex9.3, 11 We calculate (31 + 32 + 33 + โฆ + 311) separately In 31 + 32 + 33 + โฆ + 311 32/31 = 3 & 33/32 = 3 Thus, (๐๐๐๐๐๐ ๐ก๐๐๐)/(๐น๐๐๐ ๐ก ๐ก๐๐๐) = (๐โ๐๐๐ ๐ก๐๐๐)/(๐๐๐๐๐๐ ๐ก๐๐๐) i.e. common ratio is same Thus, it is a G.P. First term = a = 3 Common ratio = 3^2/3^1 = 3 Since, r > 1 โด Sn = (๐(๐^๐ โ 1))/(๐ โ 1) where Sn = sum of n terms of GP n is the number of terms a is the first term & r is the common ratio โด Sn = (๐(๐^๐ โ 1))/(๐ โ 1) Putting values a = 3 , r = 3, n = 11 S11 = (3[311โ1])/(3 โ 1) S11 =(3[311โ1])/2 Hence 31 + 32 + โฆ + 311 = (3[311โ1])/2 From (1) Putting 31 + 32 + โฆ + 311 = (3[311โ1])/2 = 22 + (3(311 โ 1))/2 =22 + 3/2(311 โ1) Therefore,
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