Principle of Mathematical Induction - Test Papers

 CBSE Test Paper 01

CH-04 Principle of Mathematical Induction 

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1. 34+1516+6364+...........$\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+...........$ to n terms is equal to

   1. n+4n3−13$n+\frac{4^n}{3}-\frac{1}{3}$

   2. n+4−n3−13$n+\frac{4^{-n}}{3}-\frac{1}{3}$

   3. n−4n3−13$n-\frac{4^n}{3}-\frac{1}{3}$

   4. n+4−n3+13$n+\frac{4^{-n}}{3}+\frac{1}{3}$

2. The greatest positive integer , which divides n ( n + 1 ) ( n + 2 ) ( n + 3 ) for all n ∈$\in$ N , is

   1. 120

   2. 6

   3. 24

   4. 2

3. For all positive integers n, the number 4n+15n−1$4^n+15n-1$ is divisible by :

   1. 16

   2. 24

   3. 9

   4. 36

4. If 49n+16n+λ${49}^n+16n+\lambda$ is divisible by 64 for all n ∈$\in$ N , then the least negative integral value of λ$\lambda$ is

   1. -1

   2. -3

   3. -4

   4. -2

5. For n∈N,xn+1+(x+1)2n−1$n\in N,x^{n+1}+{\left(x+1\right)}^{2n-1}$ is divisible by :

   1. x2+x+1$x^2+x+1$

   2. x2+x−1$x^2+x-1$

   3. x+1$x+1$

   4. x

6. Fill in the blanks:

   If a1 = 2 and an = 5 an - 1, then the value of a3 in the sequence is _______.

7. Fill in the blanks:

   If xn -1 is divisible by x - k, then the least positive integral value of k is ________.

8. Prove by the principle of mathematical induction that for all n ∈$\in$ N, 32n  when divided by 8, the remainder is always 1.

9. Prove by Mathematical Induction that the sum of first n odd natural numbers is n2.

10. Let U1 = 1, U2 = 1 and Un+2 = Un+1 + Un for n ≥$\ge$ 1. Use mathematical induction to show that:
    Un = 15√{(1+5√2)n−(1−5√2)n}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n\right\}$ for all n ≥$\ge$ 1.

CBSE Test Paper 01
CH-04 Principle of Mathematical Induction 

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Solution

1. (b) n+4−n3−13$n+\frac{4^{-n}}{3}-\frac{1}{3}$

   Explanation: When n = 1 we get 3/4$3/4$, and the subsequent terms when n is replaced by 2,3,4...

2. (c) 24
   Explanation: If n = 1 then the statement becomes 1x2x3x4= 24 : the consecutive natural numbers when substituted will be multiples of 24.
3. (c) 9
   Explanation: Replace n = 1 we get 18 n = 2 we get 45.... By the principle of mathematical induction it is divisible by 9.
4. (a) -1
   Explanation: When n = 1 we have the value of the expression as 65 . Given that the expression is divisible be 64. Hence the value is -1.
5. (a) x2+x+1$x^2+x+1$

   Explanation: When n = 1 we get x2 +$+$x +$+$1

6. 50

7. 1

8. Let P(n) be the statement given by 
   P(n) : 32n when divided by 8, the remainder is 1
   or, P(n) : 32 = 8λ$\lambda$ + 1 for some λ$\lambda$ ∈$\in$ N
   P(1): 32 = 8λ$\lambda$ + 1 for some λ$\lambda$ ∈$\in$ N.
   ∴$\therefore$ 32 = 8 ×$\times$ 1 + 1 = 8λ$\lambda$ +1, where λ$\lambda$ = 1
   P(1) is true
   Let P(m) be true. Then, 32m = 8λ$\lambda$ + 1 for some λ$\lambda$ ∈$\in$ N ...(i)
   We shall now show that P(m + 1) is true for which we have to show that 32(m + 1) when divided by 8, the remainder is 1 i.e. 32(m + 1) = 8μ$\mu$ + 1 for some μ$\mu$ ∈$\in$ N.
   Now, 32(m + 1) = 32m ×$\times$ 32 = (8λ$\lambda$ + 1) ×$\times$ 9 [Using (i)]
   = 72λ$\lambda$ + 9 = 72λ$\lambda$ + 8 + 1 = 8 (9λ$\lambda$ + 1) + 1 = 8μ$\mu$ +1, where μ$\mu$ = 9λ$\lambda$ +1 ∈$\in$ N
   ⇒$\Rightarrow$ P(m + 1) is true
   Thus, P (m) is true ⇒$\Rightarrow$ P (m + 1) is true.
   Hence, by the principle of mathematical induction P(n) is true for all n ∈$\in$ N i.e. 32n when divided by 8 the remainder is always 1.

9. Step I Let P(n) denotes the given statement, i.e.,
   P(n):1+3+5+...n(terms)=n2$P(n):1+3+5+...n(terms)=n^2$
   i.e.,P(n):1+3+5+...+(2n−1)=n2$P(n):1+3+5+...+(2n-1)=n^2$
   Since,
   First term = 2 ×$\times$ 1 - 1  = 1 
   Second term = 2 ×$\times$ 2 - 1 = 3
   Third term = 2×$\times$ 3 - 1 = 5 ......
   ∴$\therefore$ nth term=2n−1$=2n-1$
   Step II For n = 1, we have
   LHS=2.1−1=1$=2.1-1=1$
   RHS =12=1=$=1^2=1=$LHS
   Thus, P(1) is true.
   Step III For n = k, let us assume that P(k) is true,
   i.e., P(k):1+3+5+...+(2k−1)=k2$P(k):1+3+5+...+(2k-1)=k^2$ ...(i)
   Step IV For n = k + 1, we have to show that P(k + 1) is true, whenever P(k) is true i.e., 
   P(k + 1) :1+3+5+...+(2k−1)+[2(k+1)−1]=(k+1)2$1+3+5+...+(2k-1)+[2(k+1)-1]=(k+1{)}^2$
   LHS =1+3+5+...+(2k−1)+[2(k+1)−1]$=1+3+5+...+(2k-1)+[2(k+1)-1]$
   =k2+2(k+1)−1$=k^2+2(k+1)-1$ [from Eq. (i)]
   =k2+2k+1=(k+1)2=$=k^2+2k+1=(k+1{)}^2=$ RHS
   So, P(k + 1) is true, whenever, P(k) is true.
   Hence, by Principle of Mathematical Induction, P(n) is true for all n∈N$n\in N$.

10. Let P(n) be the statement given by
    P(n) : Un  =15√{(1+5√2)n−(1−5√2)n}$=\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n\right\}$
    We have,
    U1 = 15√{(1+5√2)1−(1−5√2)1}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^1-{\left(\frac{1-\sqrt{5}}{2}\right)}^1\right\}$= 1
    and,
    U2 = 15√{(1+5√2)2−(1−5√2)2}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^2-{\left(\frac{1-\sqrt{5}}{2}\right)}^2\right\}$ = 15√{(1+5+25√4)−(1+5−25√4)}$\frac{1}{\sqrt{5}}\left\{\left(\frac{1+5+2\sqrt{5}}{4}\right)-\left(\frac{1+5-2\sqrt{5}}{4}\right)\right\}$ = 1
    ∴$\therefore$ P(1) and P(2) are true.
    Let P(n) be true for all n ≤$\le$ m
    i.e. Un = 15√{(1+5√2)n−(1−5√2)n}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n\right\}$ for all n ≤$\le$ m ...(i)
    We shall now show that P(n) is true for n = m + 1.
    i.e. Um+1 = 15√{(1+5√2)m+1−(1−5√2)m+1}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^{m+1}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{m+1}\right\}$
    We have,
    Un+2 = Un+1 + Un for n ≥$\ge$ 1
    ⇒$\Rightarrow$ Um+1 = Um + Um-1 for m ≥$\ge$ 2 [On replacing n by (m-1)]
    ⇒$\Rightarrow$ Um+1 = 15√{(1+5√2)m−(1−5√2)m}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^m-{\left(\frac{1-\sqrt{5}}{2}\right)}^m\right\}$ + 15√{(1+5√2)m−1−(1−5√2)m−1}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^{m-1}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{m-1}\right\}$[Using (i)]
    ⇒$\Rightarrow$ Um+1 = 15√[{(1+5√2)m+(1+5√2)m−1}$\frac{1}{\sqrt{5}}[\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^m+{\left(\frac{1+\sqrt{5}}{2}\right)}^{m-1}\right\}$ - {(1−5√2)m+(1−5√2)m−1}$\left\{{\left(\frac{1-\sqrt{5}}{2}\right)}^m+{\left(\frac{1-\sqrt{5}}{2}\right)}^{m-1}\right\}$
    ⇒$\Rightarrow$ Um+1 = 15√{(1+5√2)m−1(1+5√2+1)−(1−5√2)m−1(1−5√2+1)}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^{m-1}\left(\frac{1+\sqrt{5}}{2}+1\right)-{\left(\frac{1-\sqrt{5}}{2}\right)}^{m-1}\left(\frac{1-\sqrt{5}}{2}+1\right)\right\}$
    ⇒$\Rightarrow$ Um+1 = 15√{(1+5√2)m−1(3+5√2)−(1−5√2)m−1(3−5√2)}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^{m-1}\left(\frac{3+\sqrt{5}}{2}\right)-{\left(\frac{1-\sqrt{5}}{2}\right)}^{m-1}\left(\frac{3-\sqrt{5}}{2}\right)\right\}$
    ⇒$\Rightarrow$ Um+1 = 15√{(1+5√2)m−1(6+25√4)−(1−5√2)m−1(6−25√4)}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^{m-1}\left(\frac{6+2\sqrt{5}}{4}\right)-{\left(\frac{1-\sqrt{5}}{2}\right)}^{m-1}\left(\frac{6-2\sqrt{5}}{4}\right)\right\}$ 
    ⇒$\Rightarrow$ Um+1 = 15√{(1+5√2)m−1(1+5√2)2−(1−5√2)m−1(1−5√2)2}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^{m-1}{\left(\frac{1+\sqrt{5}}{2}\right)}^2-{\left(\frac{1-\sqrt{5}}{2}\right)}^{m-1}{\left(\frac{1-\sqrt{5}}{2}\right)}^2\right\}$
    ⇒$\Rightarrow$ Um+1 = 15√{(1+5√2)m+1−(1−5√2)m+1}$\frac{1}{\sqrt{5}}\left\{{\left(\frac{1+\sqrt{5}}{2}\right)}^{m+1}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{m+1}\right\}$
    ∴$\therefore$ P(m + 1) is true.
    Thus, P(n) is true for all n ≤$\le$ m ⇒$\Rightarrow$ P(n) is true for all n ≤$\le$ m + 1.
    Hence, P(n) is true for all n ∈$\in$ N.