Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Binomial Theorem question

2022 · 29 Jul · Shift 1 · Q43
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Binomial Theorem
  5. /2022 · 29 Jul · Shift 1 · Q43

Binomial Theorem question

2022 · 29 Jul · Shift 1 · Q43

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (24+134)n\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}(42​+43​1​)n, in the increasing powers of 134\frac{1}{\sqrt[4]{3}}43​1​ be 64:1\sqrt[4]{6}: 146​:1. If the sixth term from the beginning is α34\frac{\alpha}{\sqrt[4]{3}}43​α​, then α\alphaα is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 84

  1. Given expansion

We have (24+134)n\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^n(42​+43​1​)n with terms written in increasing powers of 134\dfrac{1}{\sqrt[4]{3}}43​1​.

Let a=24,b=134.a=\sqrt[4]{2},\qquad b=\frac{1}{\sqrt[4]{3}}.a=42​,b=43​1​. Then the general term is Tr+1=(nr)an−rbr.T_{r+1}=\binom{n}{r}a^{n-r}b^r.Tr+1​=(rn​)an−rbr.

  1. Fifth term from the beginning

The 5th term from the beginning corresponds to r=4r=4r=4: T5=(n4)an−4b4.T_5=\binom{n}{4}a^{n-4}b^4.T5​=(4n​)an−4b4.

Since b4=(134)4=13,b^4=\left(\frac{1}{\sqrt[4]{3}}\right)^4=\frac{1}{3},b4=(43​1​)4=31​, we get T5=(n4)(24)n−4⋅13.T_5=\binom{n}{4}(\sqrt[4]{2})^{n-4}\cdot \frac{1}{3}.T5​=(4n​)(42​)n−4⋅31​.

  1. Fifth term from the end

In the expansion of (a+b)n(a+b)^n(a+b)n, the 5th term from the end is the (n−3)(n-3)(n−3)th term from the beginning, i.e. corresponds to r=n−4r=n-4r=n−4. So, T_{\text{5th from end}}=\binom{n}{n-4}a^4b^{n-4}=inom{n}{4}a^4b^{n-4}.

Since a4=2,a^4=2,a4=2, this becomes T5th from end=(n4)⋅2(134)n−4.T_{\text{5th from end}}=\binom{n}{4}\cdot 2\left(\frac{1}{\sqrt[4]{3}}\right)^{n-4}.T5th from end​=(4n​)⋅2(43​1​)n−4.

  1. Use the given ratio

Given T5:T5th from end=64:1.T_5 : T_{\text{5th from end}}=\sqrt[4]{6}:1.T5​:T5th from end​=46​:1. So, T5T5th from end=64.\frac{T_5}{T_{\text{5th from end}}}=\sqrt[4]{6}.T5th from end​T5​​=46​.

Now,

=\frac{\binom{n}{4}(\sqrt[4]{2})^{n-4}\cdot \frac13}{\binom{n}{4}\cdot 2\left(\frac{1}{\sqrt[4]{3}}\right)^{n-4}}.$$ Cancel $\binom{n}{4}$: $$=\frac{(\sqrt[4]{2})^{n-4}}{6}\cdot (\sqrt[4]{3})^{n-4}.$$ Thus, $$\frac{T_5}{T_{\text{5th from end}}}=\frac{(\sqrt[4]{6})^{n-4}}{6}.$$ Set equal to $\sqrt[4]{6}$: $$\frac{(\sqrt[4]{6})^{n-4}}{6}=\sqrt[4]{6}.$$ Since $6=(\sqrt[4]{6})^4$, we get $$ (\sqrt[4]{6})^{n-4}=(\sqrt[4]{6})^5.$$ Hence, $$n-4=5 \implies n=9.$$ 5. **Find the sixth term from the beginning** The 6th term corresponds to $r=5$: $$T_6=\binom{9}{5}a^{4}b^5.$$ Now, $$\binom{9}{5}=126, \qquad a^4=2, \qquad b^5=\left(\frac{1}{\sqrt[4]{3}}\right)^5=\frac{1}{3\sqrt[4]{3}}.$$ Therefore, $$T_6=126\cdot 2\cdot \frac{1}{3\sqrt[4]{3}} =\frac{252}{3\sqrt[4]{3}} =\frac{84}{\sqrt[4]{3}}.$$ So if the sixth term is $\dfrac{\alpha}{\sqrt[4]{3}}$, then $$\alpha=84.$$
PreviousNext

More from Binomial Theorem

  • If the constant term in the expansion of (3x3−2x2+x55​)10 is 2k.l, where l is an odd integer, then the value of k is equal to:2022 · MCQ
  • Let the coefficients of x − 1 and x − 3 in the expansion of (2x51​−x51​1​)15,x>0, be m and n respectively. If r is a positive integer such that mn2=15Cr​.2r…2022 · Numerical
  • For two positive real numbers a and b such that a21​+b31​=4, then minimum value of the constant term in the expansion of (ax81​+bx−121​)10 is :2022 · MCQ
  • If n is the number of irrational terms in the expansion of (31/4+51/8)60, then (n − 1) is divisible by :2021 · MCQ
  • If the fourth term in the expansion of (x+xlog2​x)7 is 4480, then the value of x where x ∈ N is equal to :2021 · MCQ
  • If (2021)3762 is divided by 17, then the remainder is ​.2021 · Numerical
  • Let the coefficients of third, fourth and fifth terms in the expansion of (x+x2a​)n,xe0, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to ​…2021 · Numerical
  • The term independent of x in the expansion of [x2/3−x1/3+1x+1​−x−x1/2x−1​]10, x e 1, is equal to ​.2021 · Numerical