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Home / Doubts and Solutions/KERALA/Class 11 Science/Mathematics

Find the sum to n terms of the series: 5+11+19+29+41+....
 
[I can't understand special series at all so please explain the steps as much as possible.]
 
[for example: in the answer Sn=5+11+19+29+..+an-1+an
                                     Sn=5+11+19+...+an-2+an-1+an
Shouldn't subtracting the two terms give us 0 as the result?]

Asked by Deebu 1st December 2014, 9:55 PM
Answered by Expert
Answer:
G i v e n space S equals 5 plus 11 plus 19 plus 29 plus 41 plus... plus T subscript n minus 1 end subscript plus T subscript n H e r e space w e space a r e space n o t space s u b t r a c t i n g space t h e space w h o l e space s e r i e s. O b s e r v e space t h e space f o l l o w i n g space s u b t r a c t i o n. space space space space space space space space space space space space space space space S equals 5 plus 11 plus 19 plus 29 plus 41 plus... plus T subscript n minus 1 end subscript plus T subscript n... left parenthesis 1 right parenthesis space space space space space space space space space space space space space space bottom enclose space space space space S equals minus open parentheses 5 plus 11 plus 19 plus 29 plus 41 plus... plus T subscript n minus 1 end subscript plus T subscript n close parentheses end enclose..... left parenthesis 2 right parenthesis space space space space space space space space space space space space space space space space space space 0 equals 5 plus 6 plus 8 plus 10 plus 12 plus...... plus open parentheses T subscript n minus T subscript n minus 1 end subscript close parentheses minus T subscript n space rightwards double arrow 0 equals 5 plus 6 plus 8 plus 10 plus 12 plus...... plus open parentheses T subscript n minus T subscript n minus 1 end subscript close parentheses minus T subscript n space rightwards double arrow T subscript n minus 5 equals 6 plus 8 plus 10 plus 12 plus...... plus open parentheses T subscript n minus T subscript n minus 1 end subscript close parentheses rightwards double arrow T subscript n minus 5 equals 6 plus 8 plus 10 plus 12 plus...... u p t o space n minus 1 space t e r m s rightwards double arrow T subscript n minus 5 equals S subscript n minus 1 end subscript... left parenthesis 1 right parenthesis space space space space space open square brackets S subscript n minus 1 end subscript equals 6 plus 8 plus 10 plus 12 plus...... u p t o space n minus 1 space t e r m s close square brackets S u m space o f space n space t e r m s space o f space A P space i s : S subscript n equals n over 2 open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets comma space w h e r e space a space i s space t h e space i n i t i a l space t e r m space a n d space d space i s space t h e space c o m m o n space d i f f e r e n c e. T h e r e f o r e comma space S u m space o f space open parentheses n minus 1 close parentheses space t e r m s space o f space A P space i s : S subscript n minus 1 end subscript equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 a plus open parentheses n minus 2 close parentheses d close square brackets comma space w h e r e space a space i s space t h e space i n i t i a l space t e r m space a n d space d space i s space t h e space c o m m o n space d i f f e r e n c e. W h e n space a equals 6 space a n d space d equals 2 comma space w e space h a v e comma space S subscript n minus 1 end subscript equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 cross times 6 plus open parentheses n minus 2 close parentheses cross times 2 close square brackets S u b s t i t u t i n g space t h e space v a l u e space o f space S subscript n minus 1 end subscript equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 cross times 6 plus open parentheses n minus 2 close parentheses cross times 2 close square brackets space i n space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e comma T subscript n minus 5 equals fraction numerator n minus 1 over denominator 2 end fraction open square brackets 2 cross times 6 plus open parentheses n minus 2 close parentheses cross times 2 close square brackets rightwards double arrow T subscript n minus 5 equals open parentheses n minus 1 close parentheses open square brackets 6 plus open parentheses n minus 2 close parentheses close square brackets rightwards double arrow T subscript n minus 5 equals 6 n minus 6 plus n squared minus 2 n minus n plus 2 rightwards double arrow T subscript n minus 5 equals n squared plus 3 n minus 4 rightwards double arrow T subscript n equals n squared plus 3 n plus 1 rightwards double arrow S subscript n equals sum from r equals 1 to n of open parentheses r squared plus 3 r plus 1 close parentheses rightwards double arrow S subscript n equals sum from r equals 1 to n of r squared plus sum from r equals 1 to n of 3 r plus sum from r equals 1 to n of 1 rightwards double arrow S subscript n equals sum from r equals 1 to n of r squared plus 3 sum from r equals 1 to n of r plus sum from r equals 1 to n of 1 rightwards double arrow S subscript n equals fraction numerator n open parentheses 2 n plus 1 close parentheses open parentheses n plus 1 close parentheses over denominator 6 end fraction plus fraction numerator 3 n open parentheses n plus 1 close parentheses over denominator 2 end fraction plus n
Answered by Expert 2nd December 2014, 10:20 AM
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