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Q. 1. Given that (1979) C1 + 2C2x + 3C3x2 + .............. + 2n C2nx2n – 1 = 2n (1 + x)2n – 1

where Cr =                                   r = 0, 1, 2, .................., 2n

Prove that

C1– 2C2+ 3C3– ...................... – 2nC2n2 = (– 1)nn Cn.

Ans. Sol.  Given that

C1 + 2C2x + 3C3x2 +....+ 2nC2nx2n – 1 = 2n (1 + x)2n–1    ....(1)

where

⇒ C1x + C2x2 + C3x+....+ C2nx2n = (1+ x)2n – 1
⇒ C+ C1x + C2x2 + C3x3 + ....+ C2nx2n = (1+ x)2n ....(2)

Changing x by    we get

....(3)

Multiplying eqn. (1) and (3) and equating the coefficients of x2n–1 on both sides, we get

= coeff.of x2 n-1 in 2n( x - 1) (x2- 1)2n-1

= 2n [coeff. of  x2n–2 in (x2–1) 2n–1 – coeff. of x2n–1 in (x2–1)2n–1]

⇒

(∵ 2nCn= Cn)

Hence Proved.

Q.2. Prove that 72n + (23n – 3)(3n – 1) is divisible by 25 for any natural number n. (1982 - 5 Marks)

Ans.

Sol.  P(n) : 72n + 23n –3, 3n–1 is divisible by 25 ∀n∈N .
Let us prove it by Mathematical Induction :
P(1) : 72 + 20.30 = 49 + 1= 50 which is divisible by 25.
∴ P (1) is true.
Let P(k) be true that is 72k  +23k–3, 3k–1 is divisible by 25.
⇒ 72k + 23k–3. 3k–1 = 25m where  m ∈Z .
⇒ 23k – 3 . 3k–1 = 25m – 72k ....(1)
Consider P(k + 1) : 72(k + 1) + 23(k + 1) –3 .3k + 1 – 1
= 72k .7+ 23k. 3k = 49. 72k + 23 . 3.23k–3 . 3k–1
= 49. 72k + 24 ( 25m – 72k) (Using IH eq. (1))
= 49. 72k + 24 × 25m – 24 × 72k
= 25. 72k + 24 × 25m = 25 (72k + 24 m)
= 25 × some integral value which is divisible by 25.
∴ P(k + 1) is also true.
Hence by the principle of mathematical induction
P(n) is true ∀ n∈Z .

Q.3. If (1 + x)n = C0 + C1x + C2x2 + ...... + Cnxthen show that the sum of the products of the Ci's taken two at a time,represented by   is  equal to

(1983 - 3 Marks)

Ans.

Sol.  S = ∑∑ CiC
0 ≤ i < j≤n NOTE THIS STEP

⇒ S = C0 (C1 + C+ C3 +....+ Cn) + C1 (C2 + C3 +....+ Cn) + C2 (C3 + C4 + C+.... Cn) +.... Cn–1(Cn)

⇒ S = C0 (2n – C0) + C1 (2– C– C1) + C2(2n – C0– C1– C2) + ....+ Cn – 1(2n – C0 – C1....Cn – 1) + Cn (2n – C0– C1.... Cn)

⇒ S = 2n (C0 + C+ C2 +....+ Cn – 1 + Cn)

Q.4. Use mathematical Induction to prove : If n is an y odd positive integer, then n(n2 – 1) is divisible by 24. (1983 - 2 Marks)

Ans. Sol.  P(n) : n (n2–1) is divisible by 24 for n odd +ve integer.

For n = 2m –1, it can be restated as P(m) : (2m – 1) (4m– 4m) =  4m (m – 1) (2m – 1) is divisible by 24 ∀ m ∈ N

⇒ P(m) : m (m – 1) (2m – 1) is divisible by 6 ∀ m ∈N .

Here P(1) = 0, divisible by 6.
∴ P(1) is true.

Let  it be true for m = k, i.e.,
k (k–1) (2k–1) = 6p
⇒ 2 k– 3k2 + k = 6p                ...(1)

Consider P(k + 1) : k (k + 1) (2k + 1) =2k+ 3k2 + k
= 6p + 3k2 + 3k(Using (1)
= 6 (p + k2)
⇒ divisible by 6
∴ P (k + 1) is also true.
Hence P(m) is true ∀ m ∈N .

Q.5. If p be a natural number then prove that pn + 1 + (p + 1)2n – 1 is divisible by p2 + p + 1 for every positive integer n. (1984 - 4 Marks)

Ans. Sol.  P(n) : Pn + 1 + (p + 1)2n–1 is divisible by p2 + p + 1
For n = 1, P(1) : p2 + p + 1
which is divisible  by p+ p + 1.
∴ P(1) is true.
Let P(k) be true, i.e., pk+1 + (p + 1)2k–1 is divisible by p+ p + 1
⇒ pk+1 + (p + 1)2k–1
= (p+ p + 1) m ....(1)
Consider P(k + 1) : pk+2 + (p + 1)2k+1
= p . pk+1 + (p + 1)2k–1. (p + 1)2
= p [m (p+ p + 1) – (p + 1)2k – 1] + (p + 1)2k – 1(p + 1)
= p (p2 + p + 1)m – p (p + 1)2k – 1 + (p + 1)2k – 1  (p2 + 2p + 1)
= p (p2 + p + 1)m + (p + 1)2k – 1(p2 + p + 1)
= (p2 + p + 1) [mp + (p + 1)2k – 1] = (p+ p + 1)
some integral value
∴  divisible by p2 + p + 1
∴ P (k + 1) is also true.
Hence by principle of mathematical induction P(n) is true ∀ n∈N .

Q.6. Given sn = 1 + q + q2 + ...... + qn ;

Prove that

(1984 - 4 Marks)

Ans. Sol.

We have  ....(1)

and

Now,

Using (1)

[Using eq. (2)]

Q.7. Use meth od of mathematical induction 2.7n + 3.5n – 5 is divisible by 24 for all n > 0 (1985 - 5 Marks)

Ans.

Sol.  Let An =  2.7n + 3.5n – 5
Then A1 = 2.7 + 3.5 – 5 =14 + 15 – 5 = 24.
Hence A1 is divisible by 24.
Now assume that Am is divisible by 24 so that we may write
Am = 2.7m + 3.5m - 5= 24k , k ∈N ....(1)
Then Am + 1 – A
= 2 (7m + 1 – 7m) + 3 (5m + 1 – 5m) – 5 + 5
= 2.7m(7 – 1)  + 3.5m (5 – 1) = 12. (7m + 5m)
Since 7m and 5m are odd integers ∀ m ∈ N , their sum must be an even integer, say 7m + 5m = 2p, p ∈ N .
Hence Am+1– Am=12.2 p = 24 p
or Am+1= Am + 24p = 24k + 24p [by (1)]
Hence Am+1 is divisible by 24.
It follows by mathematical induction that An is divisible by 24 for all  n ∈ N .

Q.8. Prove by mathematical induction that – (1987 - 3 Marks)

for all positive Integers n.

Ans. Sol.

Let

For n =1,

⇒    which is true for n =1

Assume that P(k) is true, then

....(1)

For n = k +1,

[Using Induction hypothesis (1)]

Thus,

In order to prove P(k + 1), it is sufficient to prove that

....(3)

Squaring eq. (3), we get

⇒ (2k + 1)2 (3k + 4) – 4 (k + 1)2 (3k + 1) ≤ 0
⇒ (4k2 + 4k + 1) (3k + 4) - 4 (k2 + 2k +1) (3k +1)≤ 0
⇒ (12k3 + 28k2 + 19k + 4) - (12k3 + 28k2 + 20k + 4)≤ 0
⇒ – k ≤ 0

which is true.
Hence from (2) and (3), we get

Hence the above inequation is true for n = k +1 and by the principle of induction it is true for all n ∈N .

Q.9. Let R =   and  f = R – [R], where [  ] denotes the greatest integer function. Prove that Rf = 42n+4 . (1988 - 5 Marks)

Ans. Sol.  We have

Therefore

This gives us  for every positive integer n.

Also

= 2k
....(1)
where k is some positive integer.

Let F

Then equation (1) becomes R – F = 2k

⇒ [R] + R – [R]– F = 2k ⇒ [R] + f – F = 2k
⇒ f – F = 2k – [R]  ⇒ f – F is an integer..
But 0 ≤ f < 1 and 0 <F<1
Therefore –1 < f – F < 1
Since f – F is an integer, we must have  f – F = 0
⇒ f  = F.
Now, Rf = RF =

Q.10. Using mathematical induction, prove that(1989 - 3 Marks)
where m, n, k are positive integers, and Cq = 0 for p < q.

Ans.

Sol.  Let the given statement be

where m, n, k ∈ N and pC= 0 for p<q.

As k is a positive integer and p Cq = 0 for p<q .
∴ k must be a positive integer less than or equal to the smaller of m and n,

We have  k = 1, when m = n = 1

1 + 1= 2.

Thus P (1, 1) is true.
Now let us assume that P(m, n) holds good for any fixed value of  m and  n  i.e.

....(2)

Consider LHS

[Using (1)]

Hence the theorem holds for the next integers  m + 1 and n +1. Then by mathematical induction the statement P (m, n) holds for all positive integral values of m and n.

Q.11. Prove that

C0 - 22C+ 32C2 - ............ + ( -1)n (n+ 1)2Cn = 0, n > 2, where Cr = nC.

Ans. Sol.  We know that (1 – x)n  =  C0 – C1x + C2x2 – C3x+....+ (– 1)n Cnx
Multiplying both sides by x, we get
x(1 – x) = C0 x – C1x2 + C2x3 – C3x4 +....+ (– 1)n Cnxn+1
Differentiating both sides w.r. to x, we get
(1 – x)n – nx (1 – x)n – 1 = C0 – 2Cx + 3Cx2 – 4C3x3 +....+ (– 1)n (n + 1) Cnx
Again multiplying both sides by x, we get
x (1 – x)n – nx(1 – x)n – 1 = C0x – 2C1x2 + 3C2x3 – 4C3x4 +....+ (– 1)n (n + 1) Cnxn + 1
Differentiating above with respect to x, we get
(1 – x)n – nx (1– x)n – 1– 2nx (1– x)n – 1 + nx2  (n – 1) (1 – x)n – 2
= C0 – 22 C1 x + 32 C2 x2 – 42 C3 x3+....+  ( – 1)n (n + 1)2 Cn x
Substituting x = 1, in above, we get
0 = C0 – 22C1+ 32C2– 42C3 +....+ ( – 1)n (n + 1)2Cn
Hence Proved.

Q.12. Prove that  is an integer for every positive integer n. (1990 -  2 Marks)

Ans. Sol.   We have

P(n) : is an integer, ∀ n∈N

P(1) :

= 1 an integer

∴ P(1) is true
Let P(k) be true i.e.

is an integer

= m, (say)

m ∈N                          ....(1)
Consider P(k + 1) :

= m + some integral value + 1
=  some integral value
∴ P (k + 1) is also true.
Hence P (n) is true ∀ n ∈N , (by the Principle of Mathematical Induction.)

Q.13. Using in duction or otherwise, prove that for any nonnegative integers m, n, r and k, (1991 -  4 Marks)

Ans. Sol.  Let

For k = 1, we will have two terms, on LHS, in sigma for m = 0 and m = 1, so that

LHS  =

and  RSH =

Hence LHS = RHS for k = 1.
Now let the formula holds for k = s, that is let

...(1)

Let us add next term corresponding to  m = s + 1 i.e.

adding to  both sides, we get

Hence the formula holds for k = s + 1 and so by the induction principle, the formula holds for all natural numbers k.

Q.14. If and ak = 1 for all k ≥ n, then show that  bn = 2n+1Cn+1 (1992 -  6 Marks)

Ans.

Sol.  Given that

....(1)

and

To prove bn=2n+1Cn+1 In the given equation (1) let us put x –3 = y so that x – 2 = y + 1 and we get

[Using ak = 1, ∀ k≥n]

NOTE THIS STEP :
nCn + n + 1C+ n + 2 Cn +....+ 2nCn = bn
⇒ (n + 1Cn + 1+ n +1Cn) + n + 2Cn +....+ 2nCn = bn
[Using nCn= n + 1Cn + 1= 1]
⇒ bn = n + 2Cn + 1 + n + 2C+....+ 2nCn
[Using mCr + mCr -–1 = m + 1Cr]
Combining the terms in similar way, we get
⇒ bn= 2nCn+1 + 2nCn ⇒ bn = 2n + 1Cn + 1
Hence Proved

Q.15. Let p ≥ 3 be an integer and α, β be the roots of x2 – (p + 1)x + 1 = 0 using mathematical induction show that α+ β.
(i) is an integer and (ii) is not divisible by p (1992 -  6 Marks)

Ans.

Sol.  Since α, β are the roots of  x2 – (p + 1) x +1 = 0
∴α + β = p + 1;  αβ = 1
Here p ≥ 3 and p ∈ Z

(i) To prove that αn + βn is an integer.
Let us consider the statement, “αn + βn is an integer.”
Then for n = 1,α + β = p + 1 which is an integer, p being an integer.
∴Statement is true for n = 1
Let the statement be true for n ≤ k, i.e., αk + βk is an integer Then ,
αk +1 + βk+1 = αk . α + βk.β
= α(αkk) + β(αk + βk) -αβkk β
= (α + β)(α k k ) -αβ(α k -1k-1
= (α +β)(α k + βk) - (α k -1 k-1) .....(1)  [as αβ = 1]
= difference of two integers = some integral value
⇒ Statement is true for n = k + 1.
∴By the principle of mathematical induction the given statement is true for ∀ n∈ N .

(ii) Let Rn be the remainder of αn when divided by p where 0 ≤ Rn ≤p-1
Since α + β = p +1     ∴R1
= 1 Also α2+ β2= (α + b)2 -2αβ = ( p + 1)2 -2
= p+ 2p – 1= p (p + 1) + p – 1
∴R2 = p –1

Also from equation (1) of previous part
(i), we have αn+1n+1 = ( p + 1) (αn) - ( αn-1 + βn-1)     =
p (αnn ) + (αn +βn ) - (αn-1n-1)
⇒ Rn+1 is the remainder of Rn – Rn–1 when divided by p
∴We  observe that R2 – R1= p – 1–1
∴R3 = p – 2
Similarly, R4 is the remainder when R3 – R2 is divided by p
where R3 – R2 = p – 2 – p + 1 = – 1 = – p + (p – 1)  ∴R4 = p – 1
R4 – R= p – 1 – p + 1=1           ∴R5= 1
R5 – R4 = 1– p + 1 = – p + 2               ∴R6 = p – 2
It is evident for above that the remainder is either 1 or p –1 or p – 2.
Since p ≥ 3, so none is divisible by p.

Q.16. Using mathematical induction, prove that tan -1(1/3) + tan -1 (1/7) + ..... tan-1{1 /(n2 +n+ 1)} = tan -1{n /(n+ 2)} (1993 -  5 Marks)

Ans. Sol.  To prove

P(n) :

For n = 1,  LHS

RHS ⇒ LHS  =  RHS.

∴P(1) is true.
Let P(k) be true, i.e.

Consider P (k + 1)

LHS  =  [Using equation (1)]

∴P(k  + 1) is also true.
Hence by the principle of mathematical induction P(n) is true for every natural number.

Q. 17. Prove that  = 0, where k = (3n) /2 and n is an even positive integer. (1993 -  5 Marks)

Ans. Sol.

To evaluate where

and n is +ve even interger.

Let n = 2m, where m ∈ z+

...(1)

Now we know that

...(2)

Keeping in mind the form of RHS in equation (1) and in equation (2)

We put  in equation (2) to get

But

NOTE THIS STEP

[Using D’ Moivre’s thm.]

Similarly,

Substituting the above in equation (3) we get

Hence Proved

Q.18. If x is not an integral multiple of 2π use mathematical induction to prove that : (1994 -  4 Marks)

Ans. Sol.  Let  P (n) : cos x + cos 2x +....+ cos nx

....(1)

where x is not an integral multiple of 2 p .
For n = 1 P (1) : L.H.S.  = cos x

R.H.S.

L.H.S. = R.H.S.
⇒ P (1) is true.

Let P(k) be true i.e.
cos x + cos 2x + ....+ cos kx

....(2)

Consider P(k + 1) :
cos x + cos 2x + ....+ cos kx + cos (k + 1) x

.L.H.S. [cos x + cos 2x + ....+ cos kx + cos (k + 1) x

[Using (2)]

= R.H.S.

∴P (k + 1) is also true.
Hence by the principle of mathematical induction
P (n) is true ∀ n∈ N .

Q.19. Let n be a positive integer and (1994 -  5 Marks)
(1 + x + x2)n = a0 + a1x + ............+ a2n x2n
Show that a02 – a12 + a22 .............+ a2n2 = an

Ans. Sol.  Given that,

(1 + x + x2)n = a0 + a1x +....+ a2nx2n ....(1)
where n is a +ve integer.
Replacing x byin eq n (1), we get

....(2)

Multiplying eq.’s (1) and (2) :

Equating the constant terms on both sides we get

constant term in the expansion of

= Coeff. of x2n in the expansion of (1 + x2 + x4)n But replacing x by x2 in eq’s (1), we have

(1 + x2 + x4)n = a+ a1x2 +....+ a2n (x2)2n
∴Coeff of x2n = a
Hence we obtain,

Q.20. Using mathematical induction prove that for every integer n ≥ 1, (32n–1) is divisible by 2n+2 but not by 2n+3. (1996 - 3 Marks)

Ans. Sol.  For n = 1, 32 n - 1=321 - 1 =9 - 1= 8 which is  divisible by 2n+2 = 23 = 8 but is not divisible by  2n+3 = 24 = 16
Therefore, the result is true for n = 1.
Assume that the result is true for n = k.
That is, assume that 32 k –1 is divisible by 2 k + 2 but is  not divisible by 2k + 3,
Since 32 k –1 is divisble by 2k + 2 but not by 2k + 3,
we can write 32 k –1= (m) 2k + 2 where m must be an odd positive integer, for otherwise 32k – 1 will become divisible by 2k + 3.

For n = k +1, we have

= (m.2k + 2 + 1)2 – 1 [Using (1)]
= m2.(2k + 2)2 + 2m.2k + 2 + 1–1
= m2.22k + 4 + m.2k + 3
= 2k + 3(m2.2k + 1 + m.)
⇒ 32k+1– 1 is divisible by 2k +3 .
But 32k +1 – 1 is not divisible by 2k + 4 for otherwise we must have 2 divides m2. 2k+1+ m.
But this is not possible as m is odd.
Thus, the result is true for  n = k + 1.

Q.21. Let 0 < Ai < p for i = 1, 2 ...., n. Use mathematical induction to prove that

sin A1 + sin A2 ... + sin An ≤ n sin

where ≥ 1 is a natural number. {You may use the fact that p sin x + (1–p) sin y ≤ sin [px + (1–p)y],   where 0 ≤ p ≤ 1 and 0 ≤ x, y ≤ π} (1997 - 5 Marks)

Ans.

Sol.  For n = 1, the inequalitity becomes sin A1 ≤ sinA1 , which is clearly true.
Assume that the inequality holds for n = k where k is some positive integer. That is, assume that

sin A1 + sin A2 + .... + sin Ak ≤ k sin ....(1)
for same positive integer k.
We shall now show that the result holds for n = k + 1 that is, we show that

sin A1 + sin A2 + .... + sin Ak+ sinAk +1

....(2)

L.H.S. of (2) =  sin A1 + sin A2 + .... + sin Ak+ sinAk +1

[Induction assumption]

where

[Using the fact p sin x + (1– p) sin y ≤ sin [px + (1– p)y]

Thus, the inequality holds for n = k + 1. Hence, by the principle of mathematical induction the inequality holds for all n∈ N.

Q.22. Let p be a prime and m a positive integer. By mathematical induction on m, or otherwise, prove that whenever r is an integer such that p does not divide r, p divides mpCr, (1998 - 8 Marks)
[Hint: You may use the fact that (1+x)(m+1)p = (1 + x)(1 + x)mp]

Ans. Sol.  We know that

Now, L.H.S is an integer
⇒ RHS must be an integer
But p and r are coprime (given)

∴r must divide m. mp-1Cr -1

oris an integer..

is an integer or mpCr is divisible by p.

Q.23. Let n be any positive integer. Prove that (1999 - 10 Marks)

for each non-be gatuve integer m ≤ n.

Ans. Sol.

Let    P(m) =

....(1)

For m = 0, LHS  =

R.H.S. = = LHS

[∵ m = 0 ⇒ k = 0]
∴P(0) holds true. Now assuming P (m)
L.H.S. of   P(m + 1) = L.H.S.  of

= R.H.S.  of   P(m + 1).
Hence by mathematical induction, result follows for all 0 ≤ m ≤ n.

Q.24. For any positive integer m, n (with n ≥ m), let

Prove that

Hence or otherwise, prove that

(2000 - 6 Marks)

Ans. Sol.  Given that for positive integers m and n such that n ≥m, then to prove that
nCm + n – 1Cm + n – 2C+....+ mC= n + 1Cm + 1
L.H.S. mCm + m + 1Cm + m + 2Cm +....+ n–1Cm + n Cm
[writing L.H.S. in reverse order]
= (m + 1Cm+1 + m + 1Cm) + m + 2Cm +....+ n–1Cm + n Cm
[∵ mCm = m + 1Cm+1]
= (m + 2Cm + 1 + m + 2Cm) + m + 3Cm +....+ nCm
[∵ nCr + 1 + nCr= n+1Cr + 1]
= m + 3Cm + 1 + m + 3Cm +....+ nCm
Combining in the same way we get
= nCm + 1 + nC= n + 1Cm + 1= R.H.S.
Again we have to prove nCm + 2 n – 1Cm +3 n – 2C+....+ (n – m + 1) mCm = n + 2Cm + 2
= [nCm + n – 1Cm + n – 2Cm +....+ mCm] + [ n – 1Cm  + n – 2Cm +....+ mCm ] + [ n – 2Cm +....+ mCm ] +....+ [ mCm]
[n – m + 1 bracketed terms] = n + 1Cm + 1 + nCm + 1 n – 1Cm+1 ....+ m + 1Cm + 1
[using previous result.]
= n + 2Cm + 2
[Replacing n by n + 1 and m by m + 1 in the previous result.] = R.H.S.

Q.25. For every positive integer n, prove that

Hence or other wise,

prove that  where [x] denotes the greatest integer not exceeding x. (2000 - 6 Marks)

Ans. Sol.  For n > 0

Now,to be proved.

I. To prove

Squaring both sides in

⇒ 4n +1< n + n +1+

which is true.

II. To prove

Squaring both sides,

n + n + 1 +

Squaring again

4 [n (n + 1)] < 4n2 + 1+ 4n or 0 < 1 which is true

Hence

Further to prove  we  have to prove that there is no positive integer which lies between

or  Using Mathematical induction.

We have to check

= 2, which is true

Assume for n = k (arbitrary)

i.e.,  To prove for n = k +1

To check   since k ≥ 0

Here 4k + 5 is an odd number and 4k + 6 is even number.
Their greatest integer will be different iff 4k + 6 is a perfect square that is 4k + 6 = r2

is not integer. But k has to be integer..
So 4k + 6 cannot be perfect square.

By Sandwich theorem

Q.26. Let a, b, c be positive real numbers such that b2 - 4ac > 0 and let α1 = c. Prove by induction that

is well - defined and

(Here, ‘well - defined’ means that the denominator in the expression for αn + 1 is not zero.) (2001 - 5 Marks)

Ans. Sol.  We have a, b, c the +ve real number s.t. b2 – 4ac > 0; α1= c.

is well defined and

For n = 1,

Now, b2 – 4ac > 0 ⇒ b– 2ac > 2ac > 0

∴  αis well defined (as denomination is not zero)

Also

∴P(n) is true for n =1.
Let the statement be true for

is well defined

and

Now, we will prove that P(k + 1) is also true

i.e.

well defined and

We have

(by IH)

Now,

∴αk + 2 is well defined. Again by IH we have

∴P(k+1) is also true.
Thus by the Principle of Mathematical Induction the Statement P(n) is true ∀ n∈ N .

Q.27. Use mathematical induction to show that (25)n+1 – 24n + 5735 is divisible by (24)2 for all n = 1, 2, ........ (2002 - 5 Marks)

Ans. Sol.  Let P(n) : (25) n + 1 – 24n + 5735 For n = 1.
P(1)   : 625  – 24 + 5735 = 6336 = (24)2 ×  (11),
which is divisible by 242.
Hence P(1) is true Let P(k) be true, where k ≥ 1
⇒ (25) k + 1 – 24k + 5735
= (24)2λ where λ ∈ N
For n = k + 1,
P (k + 1)  :  (25) k + 2 – 24 (k + 1) + 5735
= 25 [ (25) k + 1 – 24k + 5735]
+ 25.24.k – (25) (5735) + 5735 – 24 (k + 1)
= 25 (24) 2 λ + (24)2 k – 5735 × 24 – 24
= 25 (24)2 λ + (24)2 k – (24) (5736)
= 25 (24) 2 λ+ (24) 2 k – (24) 2 (239),
= (24)2 [25 λ + k – 239]
which is divisible by (24)2.
Hence, by the method of mathematical induction result is true ∀ n∈ N .

Q.28. Prove that             (2003 - 2 Marks)

Ans. Sol.  To prove that

LHS of above equation can be written as

R.H.S.  Hence Proved

Q.29. A coin has probability p of showing head when tossed. It is tossed n times. Let pdenote the probability that no two (or more) consecutive heads occur. Prove that p1=1,  p2=1–pand pn=(1– p).  pn–1 + p(1 – p) pn–2  for all n ≥ 3 .
Prove by induction on n, that  for all n ≥ 1 , where a and b are the roots of quadratic equation

x2– (1 – p) x–p (1– p)=0 and

Ans. Sol.  We have α + β = 1 – p and αβ = – p (1– p)
For n = 1, pn = p= 1

Also,

For n = 2, p2 = 1– p2

Also,

which is true for n = 2
Now let result is true for k < n where n ≥ 3.

This is true for n. Hence by principle of mathematical induction, the result holds good for all n ∈ N .

The document Subjective Type Questions: Mathematical Induction and Binomial Theorem | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE is a part of the JEE Course Maths 35 Years JEE Main & Advanced Past year Papers.
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