Binomial Theorem - Test Papers

 CBSE Test Paper 01

CH-08 Binomial Theorem

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1. Find the coefficient of  x6y3 in the expansion of (x+2y)9${\left(x+2y\right)}^9$

   1. 1008

   2. 336

   3. 84

   4. 672

2. The coefficients of xp$x^p$ and xq$x^q$ ( p, q are + ve integers) in the binomial expansion of (1+x)p+q$(1+x{)}^{p+q}$ are

   1. reciprocal to each other

   2. unequal 

   3. additive inverse of each other

   4. equal

3. (5–√+1)2n+1−(5–√−1)2n+1$(\sqrt{5}+1{)}^{2n+1}-(\sqrt{5}-1{)}^{2n+1}$is

   1. 0

   2. an even positive integer

   3. an odd positive integer

   4. not an integer

4. The term independent of x in the expansion of (x−3x2)18${\left(x-\frac{3}{x^2}\right)}^{18}$ is

   1. 36$3^6$

   2. 18C636${}^{18}{C}_6{3}^6$

   3. 18C6${}^{18}{C}_6$

   4. 18C12${}^{18}{C}_{12}$

5. The 1st three terms in the expansion of (2+x3)4$(2+\frac{x}{3}{)}^4$ are

   1. 16+32x3+24x29$16+\frac{32x}{3}+\frac{24x^2}{9}$

   2. 16+34x3+24x29$16+\frac{34x}{3}+\frac{24x^2}{9}$

   3. 16+12x+316x2$16+12x+\frac{3}{16}{x}^2$

   4. 16+3x−316x2$16+3x-\frac{3}{16}{x}^2$

6. Fill in the blanks:

   The value of 8C5${}^8{C}_5$ is ________.

7. Fill in the blanks:

   In the binomial expansion of (a + b)9, the middle terms is ________ term.

8. Find the value of 8C5.

9. Find the number of terms in expansions of (2x - 3y)9.

10. Using binomial theorem, write down the expansion: (2x + 3y)5.

11. Prove that ∑nr=03rnCr=4n$\sum _{r=0}^n{3}^{rn}{C}_r=4^n$

12. Find the coefficient of x in the expansion of (1 - 3x + 7x2) (1 - x)16.

13. Expand the given expression (2x−x2)5${\left(\frac{2}{x}-\frac{x}{2}\right)}^5$

14. If the third term in the expansion of (1x+xlog10x)5${\left(\frac{1}{x}+x^{{\log }_{10}x}\right)}^5$ is 1000, then find x.

15. Find the coefficient of x7 in (ax2+1bx)11${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$ and x-7 in (ax−1bx2)11${\left(ax-\frac{1}{b{x}^2}\right)}^{11}$ and find the relation between a and b so that these coefficients are equal.

CBSE Test Paper 01
CH-08 Binomial Theorem

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Solution

1. (d) 672
   Explanation:
   We have the general term of (x+a)n is Tr+1=nCr(x)n−rar$(x+a{)}^n\text{ is }{T}_{r+1}{=}^n{C}_r(x{)}^{n-r}{a}^r$
   Now consider (x+2y)9$(x+2y{)}^9$
   Here Tr+1=9Cr(x)9−r(2y)r=9Cr(2)r(x)9−r(y)r$T_{r+1}{=}^9{C}_r(x{)}^{9-r}(2y{)}^r{=}^9{C}_r(2{)}^r(x{)}^{9-r}(y{)}^r$
   Comparing the indices of x as well as y in x6y3$x^6{y}^3$ and in Tr+1$T_{r+1}$ , we get 
   (x)9−r(y)r2=x6y3$(x{)}^{9-r}(y{)}^{r^2}=x^6{y}^3$
   ⇒r=3$\Rightarrow r=3$
   ∴T4=T3+1=9C3(x)9−3(2y)3$\therefore T_4=T_{3+1}{=}^9{C}_3(x{)}^{9-3}(2y{)}^3$
   Hence coefficient of x6y3=8.9C3=672$x^6{y}^3=8{.}^9{C}_3=672$
2. (d) equal
   Explanation:
   We have the general term of (x+a)n is Tr+1=nCr(x)n−rar$(x+a{)}^n\text{ is }{T}_{r+1}{=}^n{C}_r(x{)}^{n-r}{a}^r$
   Now consider (1+x)p+q$(1+x{)}^{p+q}$
   Here Tr+1=p+qCr(1)(p+q)−r(x)r$T_{r+1}{=}^{p+q}{C}_r(1{)}^{(p+q)-r}(x{)}^r$
   Comparing the indices of x in xp and in Tr+1$T_{r+1}$,we get p=r
   Therefore the coefficient of xp is Tp+1=p+qCp$T_{p+1}{=}^{p+q}{C}_p$
   Now again comparing the indices of x in xq and in Tr+1$T_{r+1}$ ,we get q=r
   Therefore the coefficient of xq is Tp+1=p+qCp$T_{p+1}{=}^{p+q}{C}_p$
   But we have  p+qCp=p+qCq${}^{p+q}{C}_p{=}^{p+q}{C}_q$ [∵nCr=nCn−r]$\left[{\because }^n{C}_r{=}^n{C}_{n-r}\right]$
3. (b) an even positive integer
   Explanation:
   We have (a+b)n−(a−b)n$(a+b{)}^n-(a-b{)}^n$
   =[nC0an+nC1an−1b+nC2an−2b2+nC3an−3b3+………+nCnbn]−$=\left[\begin{array}{ccccccc}n{C}_0& a^n{+}^n{C}_1& a^{n-1}b{+}^n& C_2& a^{n-2}{b}^2{+}^n{C}_3& {a^{n-3}{b}^3+\dots \dots \dots +}^n{C}_n& b^n\end{array}\right]-$
   −[nC0an−nC1an−1b+nC2an−2b2−nC3an−3b3+....................+(−1)n.nCnbn]$-\left[{}^n{C}_0{a}^n{-}^n{C}_1{a}^{n-1}b{+}^n{C}_2{a}^{n-2}{b}^2{-}^n{C}_3{a}^{n-3}{b}^3+....................+{\left(-1\right)}^n{.}^n{C}_n{b}^n\right]$
   =2[nC1an−1b+nC3an−3b3+..............]$=2\left[{}^n{C}_1{a}^{n-1}b{+}^n{C}_3{a}^{n-3}{b}^3+..............\right]$
   Leta=5–√andb=1andn=2n+1$Leta=\sqrt{5}andb=1andn=2n+1$
   Now we get (5–√+1)2n+1−(5–√−1)2n+1$(\sqrt{5}+1{)}^{2n+1}-(\sqrt{5}-1{)}^{2n+1}$
   =2[2n+1C1(3–√)2n+2n+1C3(3–√)2n−213+2n+1C5(3–√)2n−415+…]$=2[{}^{2n+1}{C}_1(\sqrt{3}{)}^{2n}{+}^{2n+1}{C}_3(\sqrt{3}{)}^{2n-2}{1}^3{+}^{2n+1}{C}_5(\sqrt{3}{)}^{2n-4}{1}^5+\dots ]$
   =2[2n+1C1(3)n+2n+1C3(3)n−1+2n+1C5(3)n−2+..............]$=2\left[{}^{2n+1}{C}_1{\left(3\right)}^n{+}^{2n+1}{C}_3{\left(3\right)}^{n-1}{+}^{2n+1}{C}_5{\left(3\right)}^{n-2}+..............\right]$
   =2(a positive integer)
   Hence we have (5–√+1)2n−(5–√−1)2n+1$(\sqrt{5}+1{)}^{2n}-(\sqrt{5}-1{)}^{2n+1}$ is an even positive integer
4. (b) 18C636${}^{18}{C}_6{3}^6$

   Explanation:
   We have the general term of (x+a)n is Tr+1=nCr(x)n−rar$(x+a{)}^n\text{ is }{T}_{r+1}{=}^n{C}_r(x{)}^{n-r}{a}^r$
   Now consider (x−3x2)18${\left(x-\frac{3}{x^2}\right)}^{18}$
   Here Tr+1=18Cr(x)18−r(−3x2)r$T_{r+1}{=}^{18}{C}_r(x{)}^{18-r}{\left(-\frac{3}{x^2}\right)}^r$
   The term independent of x means index of  x is 0.
   Comparing the indices of x in x0$x^0$ and in Tr+1$T_{r+1}$ , we get 18−r−2r=0⇒r=183=6$18-r-2r=0\Rightarrow r=\frac{18}{3}=6$
   Therefore the required term is T6+1=18C6$T_{6+1}{=}^{18}{C}_6$
   (x)18−6(−3x2)6=18C6.36$(x{)}^{18-6}{\left(-\frac{3}{x^2}\right)}^6{=}^{18}{C}_6{.3}^6$

5. (a) 16+32x3+24x29$16+\frac{32x}{3}+\frac{24x^2}{9}$

   Explanation:
   We have (x+a)n=nC0xn+nC1xn−1a+nC2xn−2a2+nC3xn−3a3+$(x+a{)}^n{=}^n{C}_0{x}^n{+}^n{C}_1{x}^{n-1}a{+}^n{C}_2{x}^{n-2}{a}^2{+}^n{C}_3{x}^{n-3}{a}^3+$
   ........+Cnan$........+C_n{a}^n$
   Now consider (2+x3)4${\left(2+\frac{x}{3}\right)}^4$
   Here x=2,a=x3,n=4$x=2,a=\frac{x}{3},n=4$
   (2+x3)4=4C0(2)4+4C1(2)3(x3)+4C2(2)2(x3)2+4C3(2)1(x3)3+......${\left(2+\frac{x}{3}\right)}^4{=}^4{C}_0(2{)}^4{+}^4{C}_1(2{)}^3\left(\frac{x}{3}\right){+}^4{C}_2(2{)}^2{\left(\frac{x}{3}\right)}^2{+}^4{C}_3(2{)}^1{\left(\frac{x}{3}\right)}^3+......$
   =16+32x3+24x29+………$=16+\frac{32x}{3}+\frac{24x^2}{9}+\dots \dots \dots$

6. 56

7. 5th and 6th

8. 8C5 = 8!(8−5)!5!$\frac{8!}{(8-5)!5!}$ [∵$\because$ nCr = n!(n−r)!r!$\frac{n!}{(n-r)!r!}$]
   = 8!3!5!=8×7×6×5×4×3×2×13×2×1×5×4×3×2×1$\frac{8!}{3!5!}=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 5\times 4\times 3\times 2\times 1}$ = 56

9. The expansion of (x + a)n has (n + 1) terms. So, the expansion of (2x - 3y)9 has (9 + 1) = 10 terms.

10. Using formula, (a+b)n=∑nk=0nCk(an−kbk)$(a+b{)}^n=\sum _{k=0}^n{}^n{C}_k(a^{n-k}{b}^k)$
    (2x + 3y)5 = 32 x5 + 240 x4y + 720 x3y2 + 1080 x2y3 + 810 xy4 + 243 y5

11. ∑nr=0⋅Cran−rbr=(a+b)n.......(1)$\sum _{r=0}^n\cdot C_r{a}^{n-r}{b}^r=(a+b{)}^n.......(1)$
    Now, ∑nr=03rnCr=∑nr=0nCr(1)n−r.3r=(1+3)n$\sum _{r=0}^n{3}^{rn}{C}_r=\sum _{r=0}^{n}n{C}_r(1{)}^{n-r}{.3}^r=(1+3{)}^n$=(1+3)n$(1+3{)}^n$ (by (1)
    =4

12. We have, (1 - 3x + 7x2) (1 - x)16
    = (1 - 3x + 7x2) (16C0 - 16C1x + 16C2x2 + ...)
    =   (16C0 - 16C1x + 16C2x2 - ...) - 3x (16C0 - 16C1x + ...) + 7x2 (16C0 - 16C1x + ...)
    Here, the term containing x is -16C1x -3 16C0x = - 16x - 3x
    ∴$\therefore$ Coefficient of x = - 16 - 3 = - 19

13. Using binomial theorem for the expansion of (2x−x2)5${\left(\frac{2}{x}-\frac{x}{2}\right)}^5$ we have
    (2x−x2)5${\left(\frac{2}{x}-\frac{x}{2}\right)}^5$=5C0(2x)5+5C1(2x)4(−x2)${=}^5{C}_0{\left(\frac{2}{x}\right)}^5{+}^5{C}_1{\left(\frac{2}{x}\right)}^4\left(\frac{-x}{2}\right)$+5C2(2x)3(−x2)2+5C3(2x)2(−x2)3${+}^5{C}_2{\left(\frac{2}{x}\right)}^3{\left(\frac{-x}{2}\right)}^2{+}^5{C}_3{\left(\frac{2}{x}\right)}^2{\left(\frac{-x}{2}\right)}^3$
    +5C4(2x)(−x2)4+5C5(−x2)5${+}^5{C}_4\left(\frac{2}{x}\right){\left(\frac{-x}{2}\right)}^4{+}^5{C}_5{\left(\frac{-x}{2}\right)}^5$
    =32x5+5⋅16x4⋅−x2+10⋅8x3⋅x24$=\frac{32}{x^5}+5\cdot \frac{16}{x^4}\cdot \frac{-x}{2}+10\cdot \frac{8}{x^3}\cdot \frac{x^2}{4}$+10⋅4x2⋅−x38+5⋅2x⋅x416+−x532$+10\cdot \frac{4}{x^2}\cdot \frac{-x^3}{8}+5\cdot \frac{2}{x}\cdot \frac{x^4}{16}+\frac{-x^5}{32}$
    =32x5−40x3+20x−5x+58x3−x532$=\frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5x+\frac{5}{8}{x}^3-\frac{x^5}{32}$

14. We have,
    T3 = 1000
    ⇒$\Rightarrow$ T2+1 = 1000
    ⇒$\Rightarrow$ 5C2(1x)5−2(xlog10x)2${}^5{\mathrm{C}}_2{\left(\frac{1}{x}\right)}^{5-2}{\left(x^{{\log }_{10}x}\right)}^2$ = 1000
    ⇒$\Rightarrow$ 10(xlog10x)2×x−3$10{\left(x^{{\log }_{10}x}\right)}^2\times x^{-3}$ = 1000
    ⇒$\Rightarrow$ x2log10x×x−3$x^{2{\log }_{10}x}\times x^{-3}$ = 100
    ⇒$\Rightarrow$ x2log10x−3$x^{2{\log }_{10}x-3}$ = 102
    ⇒$\Rightarrow$ 2log10x−3$2{\log }_{10}x-3$ = logx102${\log }_x{10}^2$ [taking logx both sides]
    ⇒$\Rightarrow$ 2log10x−3$2{\log }_{10}x-3$ = 2 logx10${\log }_{x}10$ 
    ⇒$\Rightarrow$ 2log10x−3$2{\log }_{10}x-3$ = 2log10x$\frac{2}{{\log }_{10}x}$ [using logab = 1logba$\frac{1}{{\log }_{b}a}$]
    ⇒$\Rightarrow$ 2y−3=2y$2y-3=\frac{2}{y}$, where y = log10x
    ⇒$\Rightarrow$ 2y2 - 3y - 2 = 0
    ⇒$\Rightarrow$ (2y + 1) (y - 2) = 0
    ⇒$\Rightarrow$ y = 2 or y = −12$-\frac{1}{2}$
    ⇒$\Rightarrow$ log10x = 2 or log10x = −12$-\frac{1}{2}$ ⇒$\Rightarrow$ x = 102 = 100 or x = 10−12${10}^{-\frac{1}{2}}$

15. Suppose x7 occurs in (r + l )th term of the expansion of (ax2+1bx)11${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$
    Now, 
    Tr+1 = 11Cr (ax2)11-r (1bx)r${\left(\frac{1}{bx}\right)}^r$ = 11Cr a11-r b-r x22-3r ...(i)
    This will contain x7, if
    22 - 3r = 7 ⇒$\Rightarrow$ 3r = 15 ⇒$\Rightarrow$ r = 5.
    Putting r = 5 in (i), we obtain that
    Coefficient of x7 in the expansion of (ax2+1bx)11${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$ is 11C5 a6 b-5
    Suppose x-7 occurs in (r + 1)th term of the expansion of (ax−1bx2)11${\left(ax-\frac{1}{b{x}^2}\right)}^{11}$
    Now,
    Tr+1 = 11Cr (ax)11-r (−1bx2)r${\left(-\frac{1}{b{x}^2}\right)}^r$ = 11Cr a11-r (- 1)r b-r x11 - 3r ...(ii)
    This will contain x-7, if
    11 - 3r = - 7 ⇒$\Rightarrow$ 3r = 18 ⇒$\Rightarrow$ r = 6
    Putting r = 6 in (ii), we obtain that
    Coefficient of x-7 in the expansion of (ax−1bx2)11${\left(ax-\frac{1}{b{x}^2}\right)}^{11}$ is 11C6 a5 b-6 (-1)6
    If the coefficient of x7 in (ax2+1bx)11${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$ is equal to the coefficient of x-7 in (ax−1bx2)11${\left(ax-\frac{1}{b{x}^2}\right)}^{11}$, then
    11C5 a6 b-5 = 11C6 a5 b-6 (- 1)6 ⇒$\Rightarrow$ 11C5 ab = 11C6 ⇒$\Rightarrow$ ab = 1 [∵$\because$ 11C5 = 11C6]