Question

# Find the sum of the following series to infinity : (i) 1−13+132−133+134+....∞ (ii) 8+4√2+4+....∞ (iii) 25+352+253+354+.....∞. (iv) 10 - 9 + 8.1 - 7.29 + ...... ∞ (v) 13+152+133+154+135+156+.....∞

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Solution

## (i) S∞=1−13+132−133+..... ⇒a=1, r=−13 S∞=a1−r=11+13 S∞=34. (ii) S∞=8+4√2+4+.... ⇒a=8, r=−44√2=1√2 S∞=a1−r=81−1√2 =8√2√2−1×(√2+1)(√2+1) =8(2+√2)2−1 S∞=8(2+√2) (iii) S∞=25+252+253+254+........ =(25+353+...)+(352+354+....)S∞=S′∞+S"∞ForS′∞=a1−r=251−125=25×2524=1024S′∞=512S"∞=3251−125=325×2524=324S∞=S′∞+S"∞=1024+324+1324 (iv) This infinite G.P. has first term a = 10 and common ratio = - 910=−0.9 Thus the sum of the infinite G.P. will be : 10−9+8.9−7.29+....∞a1−r[Since |r|<1] =101−(−0.9) =101.9=10019 (v) 13+152+133+154+135+156+....∞ The G.P. can be written as follows : 13+152+133+154+135+156+.....∞ =(13+133+135+....∞)+(152+154+156+....∞)=131−132+1521−152=38+124=1024=512

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