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NCERT Solutions · Class 11 Mathematics Binomial Theorem

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EXERCISE 7.1 1–10 (part 1 of 3)

  1. Expand each of the expressions in Exercises $\displaystyle 1$ to 5.

    Exercise 1

    (1−2x)5\displaystyle (1-2 x)^{5}

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    NCERT’s answer
    $\displaystyle 1-10 x+40 x^{2}-80 x^{3}+80 x^{4}-32 x^{5}$
    Binomial expansion. By the binomial theorem, \[(a+b)^n=\sum_{r=0}^{n}{}^{n}C_r\,a^{\,n-r}b^{\,r}. \]Take \(\displaystyle a=1 \), \(\displaystyle b=-2x \), \(\displaystyle n=5 \): \[(1-2x)^5=\sum_{r=0}^{5}{}^{5}C_r\,(1)^{5-r}(-2x)^r. \]Term by term, using \(\displaystyle {}^{5}C_0,\dots,{}^{5}C_5 = 1,5,10,10,5,1 \):
    \(\displaystyle r=0:\ 1 \)
    \(\displaystyle r=1:\ 5(-2x)=-10x \)
    \(\displaystyle r=2:\ 10(-2x)^2=10(4x^2)=40x^2 \)
    \(\displaystyle r=3:\ 10(-2x)^3=10(-8x^3)=-80x^3 \)
    \(\displaystyle r=4:\ 5(-2x)^4=5(16x^4)=80x^4 \)
    \(\displaystyle r=5:\ (-2x)^5=-32x^5 \)
    Note the signs alternate, because \(\displaystyle b=-2x \) is negative.\(\displaystyle (1-2x)^5 = 1-10x+40x^2-80x^3+80x^4-32x^5 \)

  2. Exercise 2

    (2x−x2)5\displaystyle \left(\frac{2}{x}-\frac{x}{2}\right)^{5}

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    NCERT’s answer
    $\displaystyle \frac{32}{x^{5}}-\frac{40}{x^{3}}+\frac{20}{x}-5 x+\frac{5}{8} x^{3}-\frac{x^{5}}{32}$
    Binomial expansion. Apply \(\displaystyle (a+b)^n=\sum_{r=0}^{n}{}^{n}C_r\,a^{\,n-r}b^{\,r} \) with \(\displaystyle a=\dfrac{2}{x} \), \(\displaystyle b=-\dfrac{x}{2} \), \(\displaystyle n=5 \): \[\left(\frac{2}{x}-\frac{x}{2}\right)^{5}=\sum_{r=0}^{5}{}^{5}C_r\left(\frac{2}{x}\right)^{5-r}\left(-\frac{x}{2}\right)^{r}. \]The coefficients \(\displaystyle {}^{5}C_r \) are \(\displaystyle 1,5,10,10,5,1 \):
    \(\displaystyle r=0:\ \dfrac{2^5}{x^5}=\dfrac{32}{x^5} \)
    \(\displaystyle r=1:\ 5\cdot\dfrac{16}{x^4}\cdot\left(-\dfrac{x}{2}\right)=-\dfrac{40}{x^3} \)
    \(\displaystyle r=2:\ 10\cdot\dfrac{8}{x^3}\cdot\dfrac{x^2}{4}=\dfrac{20}{x} \)
    \(\displaystyle r=3:\ 10\cdot\dfrac{4}{x^2}\cdot\left(-\dfrac{x^3}{8}\right)=-5x \)
    \(\displaystyle r=4:\ 5\cdot\dfrac{2}{x}\cdot\dfrac{x^4}{16}=\dfrac{5}{8}x^3 \)
    \(\displaystyle r=5:\ -\dfrac{x^5}{32} \)
    \(\displaystyle \left(\dfrac{2}{x}-\dfrac{x}{2}\right)^{5}=\dfrac{32}{x^{5}}-\dfrac{40}{x^{3}}+\dfrac{20}{x}-5x+\dfrac{5}{8}x^{3}-\dfrac{1}{32}x^{5} \)

  3. Exercise 3

    (2x−3)6\displaystyle (2 x-3)^{6}

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    NCERT’s answer
    $\displaystyle 64 x^{6}-576 x^{5}+2160 x^{4}-4320 x^{3}+4860 x^{2}-2916 x+729$
    Binomial expansion. Apply \(\displaystyle (a+b)^n=\sum_{r=0}^{n}{}^{n}C_r\,a^{\,n-r}b^{\,r} \) with \(\displaystyle a=2x \), \(\displaystyle b=-3 \), \(\displaystyle n=6 \): \[(2x-3)^{6}=\sum_{r=0}^{6}{}^{6}C_r\,(2x)^{6-r}(-3)^{r}. \]The coefficients \(\displaystyle {}^{6}C_r \) are \(\displaystyle 1,6,15,20,15,6,1 \):
    \(\displaystyle r=0:\ 2^6x^6=64x^6 \)
    \(\displaystyle r=1:\ 6(32x^5)(-3)=-576x^5 \)
    \(\displaystyle r=2:\ 15(16x^4)(9)=2160x^4 \)
    \(\displaystyle r=3:\ 20(8x^3)(-27)=-4320x^3 \)
    \(\displaystyle r=4:\ 15(4x^2)(81)=4860x^2 \)
    \(\displaystyle r=5:\ 6(2x)(-243)=-2916x \)
    \(\displaystyle r=6:\ (-3)^6=729 \)
    \(\displaystyle (2x-3)^{6}=64x^{6}-576x^{5}+2160x^{4}-4320x^{3}+4860x^{2}-2916x+729 \)

  4. Exercise 4

    (x3+1x)5\displaystyle \left(\frac{x}{3}+\frac{1}{x}\right)^{5}

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    NCERT’s answer
    $\displaystyle \frac{x^{5}}{243}+\frac{5 x^{3}}{81}+\frac{10}{27} x+\frac{10}{9 x}+\frac{5}{3 x^{3}}+\frac{1}{x^{5}}$
    Binomial expansion. Apply \(\displaystyle (a+b)^n=\sum_{r=0}^{n}{}^{n}C_r\,a^{\,n-r}b^{\,r} \) with \(\displaystyle a=\dfrac{x}{3} \), \(\displaystyle b=\dfrac{1}{x} \), \(\displaystyle n=5 \): \[\left(\frac{x}{3}+\frac{1}{x}\right)^{5}=\sum_{r=0}^{5}{}^{5}C_r\left(\frac{x}{3}\right)^{5-r}\left(\frac{1}{x}\right)^{r}. \]With \(\displaystyle {}^{5}C_r = 1,5,10,10,5,1 \):
    \(\displaystyle r=0:\ \dfrac{x^5}{243} \)
    \(\displaystyle r=1:\ 5\cdot\dfrac{x^4}{81}\cdot\dfrac{1}{x}=\dfrac{5x^3}{81} \)
    \(\displaystyle r=2:\ 10\cdot\dfrac{x^3}{27}\cdot\dfrac{1}{x^2}=\dfrac{10x}{27} \)
    \(\displaystyle r=3:\ 10\cdot\dfrac{x^2}{9}\cdot\dfrac{1}{x^3}=\dfrac{10}{9x} \)
    \(\displaystyle r=4:\ 5\cdot\dfrac{x}{3}\cdot\dfrac{1}{x^4}=\dfrac{5}{3x^3} \)
    \(\displaystyle r=5:\ \dfrac{1}{x^5} \)
    Every term is positive here, since both \(\displaystyle a \) and \(\displaystyle b \) are positive.\(\displaystyle \left(\dfrac{x}{3}+\dfrac{1}{x}\right)^{5}=\dfrac{x^{5}}{243}+\dfrac{5x^{3}}{81}+\dfrac{10x}{27}+\dfrac{10}{9x}+\dfrac{5}{3x^{3}}+\dfrac{1}{x^{5}} \)

  5. Exercise 5

    (x+1x)6\displaystyle \left(x+\frac{1}{x}\right)^{6}

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    NCERT’s answer
    $\displaystyle x^{6}+6 x^{4}+15 x^{2}+20+\frac{15}{x^{2}}+\frac{6}{x^{4}}+\frac{1}{x^{6}}$
    Binomial expansion. Apply \(\displaystyle (a+b)^n=\sum_{r=0}^{n}{}^{n}C_r\,a^{\,n-r}b^{\,r} \) with \(\displaystyle a=x \), \(\displaystyle b=\dfrac{1}{x} \), \(\displaystyle n=6 \): \[\left(x+\frac{1}{x}\right)^{6}=\sum_{r=0}^{6}{}^{6}C_r\,x^{6-r}\left(\frac{1}{x}\right)^{r}=\sum_{r=0}^{6}{}^{6}C_r\,x^{6-2r}. \]With \(\displaystyle {}^{6}C_r = 1,6,15,20,15,6,1 \), the powers \(\displaystyle x^{6-2r} \) run \(\displaystyle x^6,x^4,x^2,x^0,x^{-2},x^{-4},x^{-6} \): \[x^{6}+6x^{4}+15x^{2}+20+\frac{15}{x^{2}}+\frac{6}{x^{4}}+\frac{1}{x^{6}}. \]The expansion is symmetric in \(\displaystyle x \) and \(\displaystyle 1/x \), as it must be, since swapping them leaves the original expression unchanged.\(\displaystyle \left(x+\dfrac{1}{x}\right)^{6}=x^{6}+6x^{4}+15x^{2}+20+\dfrac{15}{x^{2}}+\dfrac{6}{x^{4}}+\dfrac{1}{x^{6}} \)
  6. Using binomial theorem, evaluate each of the following:

    Exercise 6

    (96)3\displaystyle (96)^{3}

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    NCERT’s answer
    $\displaystyle 884736$
    Binomial theorem on a nearby round number. Write \(\displaystyle 96 \) as \(\displaystyle 100-4 \), so that the powers of \(\displaystyle 100 \) are easy to handle: \[(96)^3=(100-4)^3={}^{3}C_0(100)^3-{}^{3}C_1(100)^2(4)+{}^{3}C_2(100)(4)^2-{}^{3}C_3(4)^3. \]With \(\displaystyle {}^{3}C_r = 1,3,3,1 \): \[=1\,000\,000-3(10\,000)(4)+3(100)(16)-64 \] \[=1\,000\,000-120\,000+4\,800-64. \]Adding: \(\displaystyle 1\,000\,000-120\,000=880\,000 \); \(\displaystyle 880\,000+4\,800=884\,800 \); \(\displaystyle 884\,800-64=884\,736 \).\(\displaystyle (96)^3 = 884736 \)

  7. Exercise 7

    (102)5\displaystyle (102)^{5}

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    NCERT’s answer
    $\displaystyle 11040808032$
    Binomial theorem on a nearby round number. Write \(\displaystyle 102 = 100+2 \): \[(102)^5=(100+2)^5=\sum_{r=0}^{5}{}^{5}C_r\,(100)^{5-r}(2)^{r}. \]With \(\displaystyle {}^{5}C_r = 1,5,10,10,5,1 \): \[=(100)^5+5(100)^4(2)+10(100)^3(4)+10(100)^2(8)+5(100)(16)+32 \] \[=10\,000\,000\,000+1\,000\,000\,000+40\,000\,000+800\,000+8\,000+32. \]Adding the six terms gives \(\displaystyle 11\,040\,808\,032 \).\(\displaystyle (102)^5 = 11040808032 \)

  8. Exercise 8

    (101)4\displaystyle (101)^{4}

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    NCERT’s answer
    $\displaystyle 104060401$
    Binomial theorem on a nearby round number. Write \(\displaystyle 101 = 100+1 \): \[(101)^4=(100+1)^4=\sum_{r=0}^{4}{}^{4}C_r\,(100)^{4-r}(1)^{r}. \]With \(\displaystyle {}^{4}C_r = 1,4,6,4,1 \): \[=(100)^4+4(100)^3+6(100)^2+4(100)+1 \] \[=100\,000\,000+4\,000\,000+60\,000+400+1. \]Adding: \(\displaystyle 104\,060\,401 \).\(\displaystyle (101)^4 = 104060401 \)

  9. Exercise 9

    (99)5\displaystyle (99)^{5}

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    NCERT’s answer
    $\displaystyle 9509900499$
    Binomial theorem on a nearby round number. Write \(\displaystyle 99 = 100-1 \): \[(99)^5=(100-1)^5=\sum_{r=0}^{5}{}^{5}C_r\,(100)^{5-r}(-1)^{r}. \]Since \(\displaystyle (-1)^r \) alternates, with \(\displaystyle {}^{5}C_r = 1,5,10,10,5,1 \): \[=(100)^5-5(100)^4+10(100)^3-10(100)^2+5(100)-1 \] \[=10\,000\,000\,000-500\,000\,000+10\,000\,000-100\,000+500-1. \]Adding step by step: \(\displaystyle 10\,000\,000\,000-500\,000\,000=9\,500\,000\,000 \); \(\displaystyle +10\,000\,000=9\,510\,000\,000 \); \(\displaystyle -100\,000=9\,509\,900\,000 \); \(\displaystyle +500-1=9\,509\,900\,499 \).\(\displaystyle (99)^5 = 9509900499 \)

  10. Exercise 10

    Using Binomial Theorem, indicate which number is larger (1.1\displaystyle 1.1) 10000\displaystyle { }^{10000} or 1000.

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    NCERT’s answer
    $\displaystyle (1.1)^{10000}>1000$
    Truncating a binomial expansion to get a bound. Write \(\displaystyle 1.1 = 1+0.1 \) and expand: \[(1.1)^{10000}=(1+0.1)^{10000}={}^{10000}C_0+{}^{10000}C_1(0.1)+{}^{10000}C_2(0.1)^2+\cdots \]Every term of this expansion is positive, because both \(\displaystyle 1 \) and \(\displaystyle 0.1 \) are positive. So the sum is at least as large as its first two terms: \[(1.1)^{10000} > 1+10000\times 0.1 = 1+1000 = 1001. \]Since \(\displaystyle 1001 > 1000 \), we get \(\displaystyle (1.1)^{10000} > 1000 \).\(\displaystyle (1.1)^{10000} \) is larger than \(\displaystyle 1000 \).