1for the consecutive coeffs to be equal
7Cr=7Cr+1
7!(r)!(7-r)!=7!(r+1)!(7-r+1)!
(r+1)!(r)!=(7-r)!(7-r+1)!
r+1=17-r+1
=18-r
(r+1)(8-r)=1
8r-r2+8-r=1
r2-7r-7=0
which is giving surd roots hence there are no such values of r for which this is possible
find the consecutive terms in the binomial expansion of (3+2x)^7 whose coefficients are equal..
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17 Answers
11yupp.... thr is a mistake... in tht part onlyy..... ill explain in class sumday...!!
262vatsalya, u have made a mistake . from line 2-3. ( the 2*3 part)
1it may be ritee.... sir ka sheet mein answers nahi likha haii... just have the sumss... soo cnt sayy.. bt i guesss u r rite..:)
17Cr37-r2r=7Cr+137-(r+1)2r+1
7!*37-r*2rr!(7-r)!=7!*37-r-1*2r+1r!(7-r-1)!
1r!(7-r)!=3*2(r+1)!(7-r-1)!
(r+1)!r!=3*2(7-r)!(7-r-1)!
(r)=3*2(7-r)
r=42-6r
7r=42
r=6.
check if you ger the results from this??
1this question was in nishant's sir paper.... i mean edu ka study material.... soo i cnt say if it wrong...!!...
1that`s what i was saying....either the first qstn is wrong or if its an mcq qstn then the option will be none of these....
62yup and after tath u will get
r=7-r
so r=7/2
but r has to be an integer!
1@vatsalya .... in the second line it will be:
7! = 7!
r! (7-r)! (r+1)!(7-r-1)! [not (7-r+1)!]
1the coeff of 3 consecutive terms in the expansion of (1+x)^n are in the ratio 1:7:42. find n.
1for the first 1 i mean equate coeff. dont include x and x+1 in that :P
1do the same fr the next ques..take ratio as given..
simple but labourious
1for the first one,
write the rth term and the r+1 th term (byjust replacing r by r+1)
and equate them,
to get r....i think we'll get some fractional value for r that means no such terms exist..
and evn if u get some integer then check it by equating r+1 and r+2th term
1i think for the 1st sum some data is wrong....i may be wrong but i think so....