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NCERT Exemplar · Class 11 Mathematics Binomial Theorem

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EXERCISE 8.3 11–20 (part 2 of 4)

  1. Exercise 11

    Find the coefficient of x4\displaystyle x^4 in the expansion of (1+x+x2+x3)11\displaystyle \left(1+x+x^2+x^3\right)^{11}.

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    NCERT’s answer
    $\displaystyle 990$
    \[1+x+x^2+x^3=(1+x)(1+x^2) \]\[\left(1+x+x^2+x^3\right)^{11}=(1+x)^{11}(1+x^2)^{11} \]\(\displaystyle x^4\) comes from \(\displaystyle x^{2k}\) of \(\displaystyle (1+x^2)^{11}\) times \(\displaystyle x^{4-2k}\) of \(\displaystyle (1+x)^{11}\), \(\displaystyle k=0,1,2\).\[\sum_{k=0}^{2}\binom{11}{k}\binom{11}{4-2k}=\binom{11}{0}\binom{11}{4}+\binom{11}{1}\binom{11}{2}+\binom{11}{2}\binom{11}{0} \]\[=330+11\cdot 55+55 \]\[=330+605+55=990 \]Answer: \(\displaystyle 990\)

  2. Exercise 12

    If p\displaystyle p is a real number and if the middle term in the expansion of (p2+2)8\displaystyle \left(\frac{p}{2}+2\right)^8 is 1120\displaystyle 1120 , find p\displaystyle p.

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    NCERT’s answer
    \(\displaystyle p= \pm 2\)
    There are $\displaystyle 9$ terms, so the middle term is the 5th.\[T_5=\binom{8}{4}\left(\frac p2\right)^{4}2^{4} \]\[70\cdot\frac{p^4}{16}\cdot 16=1120 \]\[70p^4=1120 \]\[p^4=16 \]\[p=\pm 2 \quad (p \text{ real}) \]Answer: \(\displaystyle p=\pm 2\)

  3. Exercise 13

    Show that the middle term in the expansion of (x−1x)2n\displaystyle \left(x-\frac{1}{x}\right)^{2 n} is 1×3×5×…(2n−1)⌊n×(−2)n\displaystyle \frac{1 \times 3 \times 5 \times \ldots(2 n-1)}{\lfloor n} \times(-2)^n.

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    \(\displaystyle 2n+1\) terms, so the middle term is the \(\displaystyle (n+1)\)th.\[T_{n+1}=\binom{2n}{n}x^{n}\left(-\frac1x\right)^{n}=(-1)^n\binom{2n}{n} \]Split \(\displaystyle (2n)!\) into odd and even factors.\[\binom{2n}{n}=\frac{(2n)!}{n!\,n!}=\frac{[1\cdot3\cdot5\cdots(2n-1)]\,[2\cdot4\cdot6\cdots 2n]}{n!\,n!} \]\[2\cdot4\cdot6\cdots 2n=2^n\,n! \]\[\binom{2n}{n}=\frac{1\cdot3\cdot5\cdots(2n-1)}{n!}\,2^n \]\[T_{n+1}=\frac{1\cdot3\cdot5\cdots(2n-1)}{n!}\,(-2)^n \]Answer: middle term \(\displaystyle =\dfrac{1\cdot3\cdot5\cdots(2n-1)}{n!}(-2)^n\)

  4. Exercise 14

    Find n\displaystyle n in the binomial (23+133)n\displaystyle \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n if the ratio of 7th \displaystyle 7^{\text {th }} term from the beginning to the 7th \displaystyle 7^{\text {th }} term from the end is 16\displaystyle \frac{1}{6}.

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    NCERT’s answer
    \(\displaystyle n=9\)
    The 7th term from the end is the \(\displaystyle (n-5)\)th term from the beginning.\[T_7=\binom{n}{6}\left(2^{1/3}\right)^{n-6}\left(3^{-1/3}\right)^{6} \]\[T_{n-5}=\binom{n}{n-6}\left(2^{1/3}\right)^{6}\left(3^{-1/3}\right)^{n-6} \]\(\displaystyle \binom{n}{6}=\binom{n}{n-6}\) cancels.\[\frac{T_7}{T_{n-5}}=2^{\frac{n-12}{3}}\cdot 3^{\frac{n-12}{3}}=6^{\frac{n-12}{3}} \]\[6^{\frac{n-12}{3}}=\frac16=6^{-1} \]\[\frac{n-12}{3}=-1 \]\[n=9 \]Answer: \(\displaystyle n=9\)

  5. Exercise 15

    In the expansion of (x+a)n\displaystyle (x+a)^n if the sum of odd terms is denoted by O and the sum of even term by E. Then prove that
    (i)
    O2−E2=(x2−a2)n\displaystyle \mathrm{O}^2-\mathrm{E}^2=\left(x^2-a^2\right)^n
    (ii)
    4OE=(x+a)2n−(x−a)2n\displaystyle 4 \mathrm{OE}=(x+a)^{2 n}-(x-a)^{2 n}

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    \[(x+a)^n=\sum_{r=0}^{n}\binom nr x^{n-r}a^{r} \]
    Odd-numbered terms have \(\displaystyle r\) even; even-numbered terms have \(\displaystyle r\) odd. Replacing \(\displaystyle a\) by \(\displaystyle -a\) flips only the odd-\(\displaystyle r\) terms.
    \[(x+a)^n=O+E \]
    \[(x-a)^n=O-E \]
    (i)
    \[O^2-E^2=(O+E)(O-E) \]
    \[=(x+a)^n(x-a)^n=(x^2-a^2)^n \]
    (ii)
    \[4OE=(O+E)^2-(O-E)^2 \]
    \[=(x+a)^{2n}-(x-a)^{2n} \]
    Answer: \(\displaystyle O^2-E^2=(x^2-a^2)^n\) and \(\displaystyle 4OE=(x+a)^{2n}-(x-a)^{2n}\)

  6. Exercise 16

    If xp\displaystyle x^p occurs in the expansion of (x2+1x)2n\displaystyle \left(x^2+\frac{1}{x}\right)^{2 n}, prove that its coefficient is ⌊2n⌊4n−p3⌊2n+p3.\frac{\lfloor 2 n}{\left\lfloor\dfrac{4 n-p}{3}\right. \left\lfloor\dfrac{2 n+p}{3}\right.} .

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    \[T_{r+1}=\binom{2n}{r}(x^2)^{2n-r}\left(\frac1x\right)^{r}=\binom{2n}{r}x^{4n-3r} \]For \(\displaystyle x^p\):\[4n-3r=p \Rightarrow r=\frac{4n-p}{3} \]\[2n-r=2n-\frac{4n-p}{3}=\frac{2n+p}{3} \]\[\text{coefficient}=\binom{2n}{r}=\frac{(2n)!}{r!\,(2n-r)!} \]\[=\frac{(2n)!}{\left(\dfrac{4n-p}{3}\right)!\left(\dfrac{2n+p}{3}\right)!} \]Answer: \(\displaystyle \dfrac{(2n)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}\)

  7. Exercise 17

    Find the term independent of x\displaystyle x in the expansion of (1+x+2x3)(32x2−13x)9\displaystyle \left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9.

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    NCERT’s answer
    \(\displaystyle \frac{17}{54}\)
    Only the terms of the ninth power giving \(\displaystyle x^0\), \(\displaystyle x^{-1}\), \(\displaystyle x^{-3}\) matter, paired with \(\displaystyle 1\), \(\displaystyle x\), \(\displaystyle 2x^3\) respectively.\[T_{r+1}={}^9C_r\left(\tfrac32x^2\right)^{9-r}\left(-\tfrac{1}{3x}\right)^{r}={}^9C_r\left(\tfrac32\right)^{9-r}\left(-\tfrac13\right)^{r}x^{18-3r} \]\[1\cdot x^{0}:\quad 18-3r=0\Rightarrow r=6 \]\[x\cdot x^{-1}:\quad 18-3r=-1\Rightarrow r=\tfrac{19}{3}\ \text{(not an integer, no term)} \]\[2x^3\cdot x^{-3}:\quad 18-3r=-3\Rightarrow r=7 \]\[r=6:\quad {}^9C_6\left(\tfrac32\right)^3\left(\tfrac13\right)^6=84\cdot\tfrac{27}{8}\cdot\tfrac{1}{729}=\tfrac{7}{18} \]\[r=7:\quad 2\cdot{}^9C_7\left(\tfrac32\right)^2\left(-\tfrac13\right)^7=2\cdot36\cdot\tfrac94\cdot\left(-\tfrac{1}{2187}\right)=-\tfrac{2}{27} \]\[\tfrac{7}{18}-\tfrac{2}{27}=\tfrac{21-4}{54}=\tfrac{17}{54} \]Answer: \(\displaystyle \dfrac{17}{54}\)
  8. Choose the correct answer from the given options in each of the Exercises $\displaystyle 18$ to $\displaystyle 24$ (M.C.Q.).

    Exercise 18

    The total number of terms in the expansion of (x+a)100+(x−a)100\displaystyle (x+a)^{100}+(x-a)^{100} after simplification is
    (A)
    50\displaystyle 50
    (B)
    202\displaystyle 202
    (C)
    51\displaystyle 51
    (D)
    none of these

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 51\). Odd powers of \(\displaystyle a\) cancel; even powers double.\[(x+a)^{100}+(x-a)^{100}=2\left[{}^{100}C_0x^{100}+{}^{100}C_2x^{98}a^2+\cdots+{}^{100}C_{100}a^{100}\right] \]\[r=0,2,4,\dots,100\ \Rightarrow\ \tfrac{100}{2}+1=51\ \text{terms} \]

  9. Exercise 19

    Given the integers r>1,n>2\displaystyle r>1, n>2, and coefficients of (3r)th\displaystyle (3 r)^{\mathrm{th}} and (r+2)nd\displaystyle (r+2)^{\mathrm{nd}} terms in the binomial expansion of (1+x)2n\displaystyle (1+x)^{2 n} are equal, then
    (A)
    n=2r\displaystyle n=2 r
    (B)
    n=3r\displaystyle n=3 r
    (C)
    n=2r+1\displaystyle n=2 r+1
    (D)
    none of these

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    NCERT’s answer
    (A)
    (A) \(\displaystyle n=2r\). Equal coefficients need equal lower indices, or indices adding to \(\displaystyle 2n\).\[{}^{2n}C_{3r-1}={}^{2n}C_{r+1} \]\[3r-1=r+1\Rightarrow r=1\quad(\text{rejected, } r>1) \]\[(3r-1)+(r+1)=2n\Rightarrow 4r=2n\Rightarrow n=2r \]

  10. Exercise 20

    The two successive terms in the expansion of (1+x)24\displaystyle (1+x)^{24} whose coefficients are in the ratio 1\displaystyle 1:4\displaystyle 4 are
    (A)
    3rd \displaystyle 3^{\text {rd }} and 4th \displaystyle 4^{\text {th }}
    (B)
    4th \displaystyle 4^{\text {th }} and 5th \displaystyle 5^{\text {th }}
    (C)
    5th \displaystyle 5^{\text {th }} and 6th \displaystyle 6^{\text {th }}
    (D)
    6th \displaystyle 6^{\text {th }} and 7th \displaystyle 7^{\text {th }}
    [Hint: 24Cr24Cr+1=14r+124−r14⇒4r+4=24−4⇒r=4\displaystyle \frac{{ }^{24} \mathrm{C}_r}{{ }^{24} \mathrm{C}_{r+1}}=\frac{1}{4} \quad \frac{r+1}{24-r} \quad \frac{1}{4} \Rightarrow 4 r+4=24-4 \Rightarrow \boxed{r=4} ]

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 5^{\text{th}}\) and \(\displaystyle 6^{\text{th}}\). Take the terms \(\displaystyle T_{r+1}\) and \(\displaystyle T_{r+2}\).\[\frac{{}^{24}C_r}{{}^{24}C_{r+1}}=\frac{r+1}{24-r}=\frac14 \]\[4r+4=24-r\Rightarrow r=4 \]\[T_5,\ T_6:\quad \frac{{}^{24}C_4}{{}^{24}C_5}=\frac{10626}{42504}=\frac14 \]