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Binomial Theorem question

2005 · Shift 0 · Q101
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  5. /2005 · Shift 0 · Q101

Binomial Theorem question

2005 · Shift 0 · Q101

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the coefficients of rth, (r+1)th, and (r + 2)th terms in the binomial expansion of (1+y)m{{\rm{(1 + y )}}^m}(1+y)m are in A.P., then m and r satisfy the equation
  1. A
    m2−m(4r−1)+4 r2−2=0{m^2} - m(4r - 1) + 4\,{r^2} - 2 = 0m2−m(4r−1)+4r2−2=0
  2. B
    m2−m(4r+1)+4 r2+2=0{m^2} - m(4r + 1) + 4\,{r^2} + 2 = 0m2−m(4r+1)+4r2+2=0
  3. C
    m2−m(4r+1)+4 r2−2=0{m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0m2−m(4r+1)+4r2−2=0
  4. D
    m2−m(4r−1)+4 r2+2=0{m^2} - m(4r - 1) + 4\,{r^2} + 2 = 0m2−m(4r−1)+4r2+2=0
View written solutionFree

Correct answer: C

  1. In the expansion of
(1+y)m=∑k=0m(mk)yk,(1+y)^m= \sum_{k=0}^m \binom{m}{k} y^k,(1+y)m=k=0∑m​(km​)yk,

the coefficients of consecutive terms are binomial coefficients.

  1. The rrrth, (r+1)(r+1)(r+1)th, and (r+2)(r+2)(r+2)th terms are:
  • rrrth term: coefficient (mr−1)\binom{m}{r-1}(r−1m​)
  • (r+1)(r+1)(r+1)th term: coefficient (mr)\binom{m}{r}(rm​)
  • (r+2)(r+2)(r+2)th term: coefficient (mr+1)\binom{m}{r+1}(r+1m​)

Since these coefficients are in A.P.,

2(mr)=(mr−1)+(mr+1).2\binom{m}{r}=\binom{m}{r-1}+\binom{m}{r+1}.2(rm​)=(r−1m​)+(r+1m​).
  1. Express adjacent binomial coefficients in terms of (mr)\binom{m}{r}(rm​):
(mr−1)=(mr)⋅rm−r+1,\binom{m}{r-1}=\binom{m}{r}\cdot \frac{r}{m-r+1},(r−1m​)=(rm​)⋅m−r+1r​, (mr+1)=(mr)⋅m−rr+1.\binom{m}{r+1}=\binom{m}{r}\cdot \frac{m-r}{r+1}.(r+1m​)=(rm​)⋅r+1m−r​.

Substitute into the A.P. condition:

2(mr)=(mr)⋅rm−r+1+(mr)⋅m−rr+1.2\binom{m}{r}=\binom{m}{r}\cdot \frac{r}{m-r+1}+\binom{m}{r}\cdot \frac{m-r}{r+1}.2(rm​)=(rm​)⋅m−r+1r​+(rm​)⋅r+1m−r​.

Cancelling (mr)≠0\binom{m}{r}\neq 0(rm​)=0,

2=rm−r+1+m−rr+1.2=\frac{r}{m-r+1}+\frac{m-r}{r+1}.2=m−r+1r​+r+1m−r​.
  1. Take LCM:
2=r(r+1)+(m−r)(m−r+1)(m−r+1)(r+1).2=\frac{r(r+1)+(m-r)(m-r+1)}{(m-r+1)(r+1)}.2=(m−r+1)(r+1)r(r+1)+(m−r)(m−r+1)​.

So,

2(m−r+1)(r+1)=r(r+1)+(m−r)(m−r+1).2(m-r+1)(r+1)=r(r+1)+(m-r)(m-r+1).2(m−r+1)(r+1)=r(r+1)+(m−r)(m−r+1).
  1. Expand both sides.

Left side:

(m−r+1)(r+1)=mr−r2+m+1,(m-r+1)(r+1)=mr-r^2+m+1,(m−r+1)(r+1)=mr−r2+m+1,

so

2(m−r+1)(r+1)=2mr−2r2+2m+2.2(m-r+1)(r+1)=2mr-2r^2+2m+2.2(m−r+1)(r+1)=2mr−2r2+2m+2.

Right side:

r(r+1)=r2+r,r(r+1)=r^2+r,r(r+1)=r2+r, (m−r)(m−r+1)=(m−r)2+(m−r)=m2−2mr+r2+m−r.(m-r)(m-r+1)=(m-r)^2+(m-r)=m^2-2mr+r^2+m-r.(m−r)(m−r+1)=(m−r)2+(m−r)=m2−2mr+r2+m−r.

Hence,

r(r+1)+(m−r)(m−r+1)=m2−2mr+2r2+m.r(r+1)+(m-r)(m-r+1)=m^2-2mr+2r^2+m.r(r+1)+(m−r)(m−r+1)=m2−2mr+2r2+m.

Thus,

2mr−2r2+2m+2=m2−2mr+2r2+m.2mr-2r^2+2m+2=m^2-2mr+2r^2+m.2mr−2r2+2m+2=m2−2mr+2r2+m.

Rearranging,

m2−4mr+4r2−m−2=0.m^2-4mr+4r^2-m-2=0.m2−4mr+4r2−m−2=0.

That is,

m2−m(4r+1)+4r2−2=0.m^2-m(4r+1)+4r^2-2=0.m2−m(4r+1)+4r2−2=0.
  1. Compare with options:
  • A: m2−m(4r−1)+4r2−2m^2-m(4r-1)+4r^2-2m2−m(4r−1)+4r2−2
  • B: m2−m(4r+1)+4r2+2m^2-m(4r+1)+4r^2+2m2−m(4r+1)+4r2+2
  • C: m2−m(4r+1)+4r2−2m^2-m(4r+1)+4r^2-2m2−m(4r+1)+4r2−2
  • D: m2−m(4r−1)+4r2+2m^2-m(4r-1)+4r^2+2m2−m(4r−1)+4r2+2

So the correct option is:

C\boxed{\text{C}}C​
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