Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Live 1-1 coding classes to unleash the creator in your Child
In the binomial expansion of (a - b)n, n = 5, the sum of 5th and 6th terms is 0, then a / b equals
t5+ t6= 0nC4an - 4(- b)4+nC5an - 5(- b)5= 0Solving above equation, we get a / b = (n - 4) / 5ThanksBharat Bajajaskiitians faculty
by using easy methodtr+1/tr = n-r+1/r=>n-5+1/5=n-4/5 for r =5it is a easy method which can be used to solve any iit problem
A/C to question ;t5 + t6 = 0 ——— (1)Since, t(r+1)= nCr. a^(n-r). b^rTherefore, from (1) ;nC4. a^(n-4). (-b)^4 + nC5. a^(n-5). (-b)^5=0 nC4. a^(n-4). b^4=
Sorry for incomplete answernC4. a^(n-4). b^4= nC5. a^(n-5). b^5[a^(n-4). b^4] / [a^(n-5). b^5]= nC5/nC4a^(n-4-n+5). b^(4-5)= nC5/nC4So, a/b= nC5/ nC4a/b = n!/(n-5)!4! x (n-4)!4!/n!Therefore, " a/b= (n-4)/5. "
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -