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Sir pls solve this question.

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Asked by rsudipto 14th December 2018, 9:52 AM
Answered by Expert
Answer:
(r+1)th term of binomial expansion (3 - 5x )15 is begin mathsize 12px style C presuperscript 15 subscript r space 3 to the power of 15 minus r end exponent space space open parentheses 5 x close parentheses to the power of r end style
if x= 1/5, then (r+1)th term begin mathsize 12px style C presuperscript 15 subscript r space 3 to the power of 15 minus r end exponent space space open parentheses 1 close parentheses to the power of r space equals space C presuperscript 15 subscript r space 3 to the power of 15 minus r end exponent space end style
In the given binomial expansion, there are 16 terms. Hence  nCr values  of 8th term and 9th term are greater than any other term.
But for 9th term power of 3 is less. Hence we start from 8th term and compare each preceding term to get the highest term numerically.
 
To compare nCr values of neighbouring terms, we use the following relation
begin mathsize 12px style fraction numerator C presuperscript n subscript r minus 1 end subscript over denominator C presuperscript n subscript r end fraction space equals space fraction numerator open parentheses n minus r close parentheses begin display style factorial end style begin display style space end style begin display style r end style begin display style factorial end style over denominator open parentheses n minus r plus 1 close parentheses begin display style factorial end style begin display style space end style begin display style open parentheses r minus 1 close parentheses end style begin display style factorial end style end fraction space equals space fraction numerator r over denominator open parentheses n minus r plus 1 close parentheses end fraction end style
8th term = t8 = begin mathsize 12px style C presuperscript 15 subscript 7 space space 3 to the power of 8 end style
7th term = t7 = begin mathsize 12px style C presuperscript 15 subscript 6 space space 3 to the power of 9 space space space equals space 7 over 9 cross times 3 cross times t subscript 8 space equals space 7 over 3 cross times t subscript 8 end style
6th term = t6 = begin mathsize 12px style C presuperscript 15 subscript 5 space space 3 to the power of 10 space space space equals space 6 over 10 cross times 3 cross times t subscript 7 space equals space 9 over 5 cross times t subscript 7 end style
5th term = t5 = begin mathsize 12px style C presuperscript 15 subscript 4 space space 3 to the power of 11 space space space equals space 5 over 11 cross times 3 cross times t subscript 6 space equals space 15 over 11 cross times t subscript 6 end style
4th term = t4 = begin mathsize 12px style C presuperscript 15 subscript 3 space space 3 to the power of 12 space space space equals space 4 over 12 cross times 3 cross times t subscript 5 space equals space t subscript 5 end style
3rd term = t3 = begin mathsize 12px style C presuperscript 15 subscript 2 space space 3 to the power of 13 space space space equals space 3 over 13 cross times 3 cross times t subscript 4 space equals space 9 over 13 cross times t subscript 4 end style
Hence 5th term and 4th term are numerically higher terms than any other term in the given binomial expansion
Answered by Expert 14th December 2018, 2:51 PM
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