Q. 1. Given that (1979) C_{1} +^{ 2}C_{2}x + ^{3}C_{3}x^{2} + .............. + 2n C_{2n}x^{2n – 1} = 2n (1 + x)^{2n – 1}
where Cr = r = 0, 1, 2, .................., 2n
Prove that
C_{1}^{2 }– 2C_{2}^{2 }+ 3C_{3}^{2 }– ...................... – 2nC_{2n}^{2} = (– 1)^{n}n C_{n}.
Ans. Sol. Given that
C_{1} + 2C_{2}x + 3C_{3}x^{2} +....+ 2nC_{2n}x^{2n} ^{– 1} = 2n (1 + x)^{2n–1} ....(1)
where
⇒ C_{1}x + C_{2}x^{2} + C_{3}x^{3 }+....+ C_{2n}x^{2n} = (1+ x)^{2n – 1 }
⇒ C_{0 }+ C_{1}x + C_{2}x^{2} + C_{3}x^{3} + ....+ C_{2n}x^{2n} = (1+ x)^{2n} ....(2)
Changing x by we get
....(3)
Multiplying eqn. (1) and (3) and equating the coefficients of x^{2n–1} on both sides, we get
= coeff.of x^{2 n1} in 2n( x  1) (x^{2} 1)^{2n1 }
= 2n [coeff. of x^{2n–2} in (x^{2}–1) ^{2n–1} – coeff. of x^{2n–1} in (x^{2}–1)^{2n–1}]
⇒
(∵ ^{2n}C_{n}= C_{n})
Hence Proved.
Q.2. Prove that 7^{2n} + (2^{3n – 3})(3^{n – 1}) is divisible by 25 for any natural number n. (1982  5 Marks)
Ans.
Sol. P(n) : 7^{2n} + 2^{3n –3}, 3^{n–1} is divisible by 25 ∀n∈N .
Let us prove it by Mathematical Induction :
P(1) : 7^{2} + 2^{0}.3^{0} = 49 + 1= 50 which is divisible by 25.
∴ P (1) is true.
Let P(k) be true that is 7^{2k} +2^{3k–3}, 3^{k–1} is divisible by 25.
⇒ 7^{2k} + 2^{3k–3}. 3^{k–1} = 25m where m ∈Z .
⇒ 2^{3k – 3} . 3^{k–1 }= 25m – 72k ....(1)
Consider P(k + 1) : 7^{2(k + 1)} + 2^{3(k + 1)} –3 .3^{k + 1 – 1 }
= 7^{2k} .7^{2 }+ 2^{3k}. 3^{k} = 49. 7^{2k} + 2^{3} . 3.2^{3k–3} . 3^{k–1 }
= 49. 7^{2k} + 24 ( 25m – 7^{2k}) (Using IH eq. (1))
= 49. 7^{2k }+ 24 × 25m – 24 × 7^{2k }
= 25. 7^{2k }+ 24 × 25m = 25 (7^{2k} + 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} = C_{0} + C_{1}x + C_{2}x^{2} + ...... + C_{n}x^{n }then show that the sum of the products of the C_{i}'^{s} taken two at a time,represented by is equal to
(1983  3 Marks)
Ans.
Sol. S = ∑∑ C_{i}C_{j }
0 ≤ i < j≤n NOTE THIS STEP
⇒ S = C_{0} (C_{1} + C_{2 }+ C_{3} +....+ C_{n}) + C_{1} (C_{2} + C_{3} +....+ C_{n}) + C_{2} (C_{3} + C_{4} + C_{5 }+.... C_{n}) +.... C_{n–1}(C_{n})
⇒ S = C_{0} (2^{n} – C_{0}) + C_{1} (2^{n }– C_{0 }– C_{1}) + C_{2}(2^{n} – C_{0}– C_{1}– C_{2}) + ....+ C_{n – 1}(2^{n} – C_{0} – C_{1}....C_{n – 1)} + C_{n} (2^{n} – C_{0}– C_{1}.... C_{n})
⇒ S = 2^{n} (C_{0} + C_{1 }+ C_{2} +....+ C_{n – 1} + C_{n})
Q.4. Use mathematical Induction to prove : If n is an y odd positive integer, then n(n^{2} – 1) is divisible by 24. (1983  2 Marks)
Ans. Sol. P(n) : n (n^{2}–1) is divisible by 24 for n odd +ve integer.
For n = 2m –1, it can be restated as P(m) : (2m – 1) (4m^{2 }– 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^{3 }– 3k^{2} + k = 6p ...(1)
Consider P(k + 1) : k (k + 1) (2k + 1) =2k^{3 }+ 3k^{2} + k
= 6p + 3k^{2} + 3k^{2 }(Using (1)
= 6 (p + k^{2})
⇒ 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 p^{n + 1} + (p + 1)^{2n – 1} is divisible by p^{2} + p + 1 for every positive integer n. (1984  4 Marks)
Ans. Sol. P(n) : P^{n + 1} + (p + 1)^{2n–1} is divisible by p^{2} + p + 1
For n = 1, P(1) : p^{2} + p + 1
which is divisible by p^{2 }+ p + 1.
∴ P(1) is true.
Let P(k) be true, i.e., p^{k+1} + (p + 1)2^{k–1} is divisible by p^{2 }+ p + 1
⇒ p^{k+1} + (p + 1)^{2k–1 }
= (p^{2 }+ p + 1) m ....(1)
Consider P(k + 1) : p^{k+2} + (p + 1)^{2k+1}
= p . p^{k+1} + (p + 1)^{2k–1}. (p + 1)^{2}
= p [m (p^{2 }+ p + 1) – (p + 1)^{2k – 1}] + (p + 1)^{2k – 1}(p + 1)^{2 }
= p (p^{2} + p + 1)m – p (p + 1)^{2k – 1} + (p + 1)^{2k – 1} (p^{2} + 2p + 1)
= p (p^{2} + p + 1)m + (p + 1)^{2k – 1}(p^{2} + p + 1)
= (p^{2} + p + 1) [mp + (p + 1)^{2k – 1}] = (p^{2 }+ p + 1)
some integral value
∴ divisible by p^{2} + p + 1
∴ P (k + 1) is also true.
Hence by principle of mathematical induction P(n) is true ∀ n∈N .
Q.6. Given s_{n} = 1 + q + q^{2} + ...... + q^{n} ;
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 2.7^{n} + 3.5^{n} – 5 is divisible by 24 for all n > 0 (1985  5 Marks)
Ans.
Sol. Let A_{n} = 2.7^{n} + 3.5^{n} – 5
Then A_{1} = 2.7 + 3.5 – 5 =14 + 15 – 5 = 24.
Hence A_{1} is divisible by 24.
Now assume that Am is divisible by 24 so that we may write
A_{m} = 2.7^{m} + 3.5^{m}  5= 24k , k ∈N ....(1)
Then A_{m + 1} – A_{m }
= 2 (7^{m + 1} – 7^{m}) + 3 (5^{m + 1} – 5^{m}) – 5 + 5
= 2.7^{m}(7 – 1) + 3.5^{m} (5 – 1) = 12. (7^{m} + 5^{m})
Since 7^{m} and 5^{m} are odd integers ∀ m ∈ N , their sum must be an even integer, say 7^{m} + 5^{m} = 2p, p ∈ N .
Hence A_{m+1}– A_{m}=12.2 p = 24 p
or A_{m+1}= A_{m} + 24p = 24k + 24p [by (1)]
Hence A_{m+1} is divisible by 24.
It follows by mathematical induction that A_{n} 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
⇒ (4k^{2} + 4k + 1) (3k + 4)  4 (k^{2} + 2k +1) (3k +1)≤ 0
⇒ (12k^{3} + 28k^{2} + 19k + 4)  (12k^{3} + 28k^{2} + 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 = 4^{2n+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 ^{p }C_{q} = 0 for p < q.
Ans.
Sol. Let the given statement be
where m, n, k ∈ N and ^{p}C_{q }= 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
C_{0}  2^{2}C_{1 }+ 3^{2}C_{2}  ............ + ( 1)^{n} (n+ 1)^{2}C_{n} = 0, n > 2, where C_{r} = ^{n}C_{r }.
Ans. Sol. We know that (1 – x)^{n} = C_{0} – C_{1}x + C_{2}x^{2} – C_{3}x^{3 }+....+ (– 1)^{n} C_{n}x^{n }
Multiplying both sides by x, we get
x(1 – x)^{n } = C_{0} x – C_{1}x^{2} + C_{2}x^{3} – C_{3}x^{4} +....+ (– 1)^{n} C_{n}x^{n+1}
Differentiating both sides w.r. to x, we get
(1 – x)^{n} – nx (1 – x)^{n} – 1 = C_{0} – 2C_{1 }x + 3C_{2 }x^{2} – 4C_{3}x^{3} +....+ (– 1)^{n} (n + 1) C_{n}x^{n }
Again multiplying both sides by x, we get
x (1 – x)^{n} – nx^{2 }(1 – x)n – 1 = C_{0}x – 2C_{1}x^{2} + 3C_{2}x^{3} – 4C_{3}x^{4} +....+ (– 1)^{n} (n + 1) C_{n}x^{n + 1}
Differentiating above with respect to x, we get
(1 – x)^{n} – nx (1– x)^{n} ^{– 1}– 2nx (1– x)^{n – 1} + nx^{2} (n – 1) (1 – x)^{n – 2 }
= C_{0} – 2^{2} C_{1} x + 32 C_{2} x^{2} – 4^{2} C_{3} x^{3}+....+ ( – 1)^{n} (n + 1)^{2} C_{n} x^{n }
Substituting x = 1, in above, we get
0 = C_{0} – 2^{2}C_{1}+ 3^{2}C_{2}– 4^{2}C_{3} +....+ ( – 1)^{n} (n + 1)^{2}C_{n}
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 b_{n} = ^{2n+1}C_{n+1} (1992  6 Marks)
Ans.
Sol. Given that
....(1)
and
To prove b_{n}=^{2n+1}C_{n+1} In the given equation (1) let us put x –3 = y so that x – 2 = y + 1 and we get
[Using a_{k} = 1, ∀ k≥n]
NOTE THIS STEP :
⇒ ^{n}C_{n} + ^{n + 1}C_{n }+ ^{n + 2} C_{n} +....+ ^{2n}C_{n} = b_{n}
⇒ (^{n + 1}C_{n + 1}+ ^{n +1}C_{n}) + ^{n + 2}C_{n} +....+ ^{2n}C_{n} = b_{n}
[Using ^{n}C_{n}= ^{n + 1}C_{n + 1}= 1]
⇒ b_{n} = ^{n + 2}C_{n + 1 }+ ^{n + 2}C_{n }+....+ ^{2n}C_{n}
[Using ^{m}C_{r} + ^{m}C_{r –1 }= ^{m + 1}C_{r}]
Combining the terms in similar way, we get
⇒ b_{n}= ^{2n}C_{n+1} + ^{2n}C_{n} ⇒ b_{n} = ^{2n + 1}C_{n + 1}
Hence Proved
Q.15. Let p ≥ 3 be an integer and α, β be the roots of x^{2} – (p + 1)x + 1 = 0 using mathematical induction show that α^{n }+ β^{n }.
(i) is an integer and (ii) is not divisible by p (1992  6 Marks)
Ans.
Sol. Since α, β are the roots of x^{2} – (p + 1) x +1 = 0
∴α + β = p + 1; αβ = 1
Here p ≥ 3 and p ∈ Z
(i) To prove that α^{n} + β^{n} is an integer.
Let us consider the statement, “α^{n} + β^{n} 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., α^{k} + β^{k} is an integer Then ,
α^{k +1} + β^{k+1} = α^{k} . α + β^{k.}β
= α(α^{k} +β^{k}) + β(α^{k} + β^{k}) αβ^{k} α ^{k} β
= (α + β)(α^{ k }+β^{k} ) αβ(α^{ k 1} +β^{k1}
= (α +β)(α ^{k} + β^{k})  (α ^{k 1 }+β^{k1}) .....(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 R_{n} be the remainder of α^{n }+β^{n} when divided by p where 0 ≤ R_{n} ≤p1
Since α + β = p +1 ∴R_{1}
= 1 Also α^{2}+ β^{2}= (α + b)^{2} 2αβ = ( p + 1)^{2} 2
= p^{2 }+ 2p – 1= p (p + 1) + p – 1
∴R_{2} = p –1
Also from equation (1) of previous part
(i), we have α^{n+1} +β^{n+1} = ( p + 1) (α^{n} +β^{n })  ( α^{n1} + β^{n1}) =
p (α^{n} +β^{n} ) + (αn +βn )  (α^{n1} +β^{n1})
⇒ R_{n+1} is the remainder of R_{n} – R_{n–1} when divided by p
∴We observe that R_{2} – R_{1}= p – 1–1
∴R_{3} = p – 2
Similarly, R_{4} is the remainder when R_{3} – R_{2} is divided by p
where R_{3} – R_{2} = p – 2 – p + 1 = – 1 = – p + (p – 1) ∴R_{4} = p – 1
R_{4} – R_{3 }= p – 1 – p + 1=1 ∴R_{5}= 1
R_{5} – R_{4} = 1– p + 1 = – p + 2 ∴R_{6} = 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 /(n^{2} +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 + x^{2})^{n} = a_{0} + a_{1}x + ............+ a_{2n} x^{2n }
Show that a_{0}^{2} – a_{1}^{2} + a_{2}^{2} .............+ a_{2n}^{2} = a_{n}
Ans. Sol. Given that,
(1 + x + x^{2})^{n} = a_{0} + a_{1}x +....+ a_{2n}x^{2n }....(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 x^{2n} in the expansion of (1 + x^{2} + x^{4})^{n} But replacing x by x^{2} in eq’s (1), we have
(1 + x^{2} + x^{4})^{n} = a_{0 }+ a_{1}x^{2} +....+ a_{2n} (x^{2})^{2n}
∴Coeff of x^{2n} = a_{n }
Hence we obtain,
Q.20. Using mathematical induction prove that for every integer n ≥ 1, (3^{2n}–1) is divisible by 2^{n+2} but not by 2^{n+3}. (1996  3 Marks)
Ans. Sol. For n = 1, 3^{2 n}  1=3^{21}  1 =9  1= 8 which is divisible by 2^{n+2 }= 2^{3} = 8 but is not divisible by 2^{n+3} = 2^{4} = 16
Therefore, the result is true for n = 1.
Assume that the result is true for n = k.
That is, assume that 3^{2 k} –1 is divisible by 2 ^{k + 2} but is not divisible by 2^{k + 3},
Since 3^{2 k} –1 is divisble by 2^{k + 2} but not by 2^{k + 3},
we can write 3^{2 k} –1= (m) 2^{k + 2} where m must be an odd positive integer, for otherwise 3^{2k }– 1 will become divisible by 2^{k + 3}.
For n = k +1, we have
= (m.2^{k + 2} + 1)^{2} – 1 [Using (1)]
= m^{2}.(2^{k + 2})^{2} + 2m.2^{k + 2 }+ 1–1
= m^{2}.2^{2k} ^{+ 4} + m.2^{k + 3 }
= 2^{k + 3}(m^{2}.^{2k + 1 }+ m.)
⇒ 3^{2k+1}– 1 is divisible by 2^{k +3} .
But 3^{2k +1} – 1 is not divisible by 2^{k + 4} for otherwise we must have 2 divides m^{2}. 2^{k+1}+ m.
But this is not possible as m is odd.
Thus, the result is true for n = k + 1.
Q.21. Let 0 < A_{i} < p for i = 1, 2 ...., n. Use mathematical induction to prove that
sin A_{1} + sin A_{2} ... + sin A_{n} ≤ 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 A_{1} ≤ sinA_{1} , which is clearly true.
Assume that the inequality holds for n = k where k is some positive integer. That is, assume that
sin A_{1} + sin A_{2} + .... + sin A_{k} ≤ 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 A_{1} + sin A_{2} + .... + sin A_{k}+ sinA_{k +1}
....(2)
L.H.S. of (2) = sin A_{1} + sin A_{2} + .... + sin A_{k}+ sinA_{k +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 ^{mp}C_{r}, (1998  8 Marks)
[Hint: You may use the fact that (1+x)^{(m+1)}p = (1 + x)^{p }(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. ^{mp1}C_{r 1}
oris an integer..
⇒is an integer or ^{mp}C_{r} is divisible by p.
Q.23. Let n be any positive integer. Prove that (1999  10 Marks)
for each nonbe 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
^{n}C_{m} + ^{n – 1}C_{m} + ^{n – 2}C_{m }+....+ ^{m}C_{m }= ^{n + 1}C_{m + 1 }
L.H.S. ^{m}C_{m} + ^{m + 1}C_{m} + ^{m + 2}C_{m} +....+ ^{n–1}C_{m} + ^{n} C_{m}
[writing L.H.S. in reverse order]
= (^{m + 1}C_{m+1} + ^{m + 1}C_{m}) +^{ m + 2}C_{m} +....+ ^{n–1}C_{m} +^{ n }C_{m}
[∵ ^{m}C_{m} = ^{m + 1}C_{m+1}]
= (^{m + 2}C_{m + 1} + ^{m + 2}C_{m}) + ^{m + 3}C_{m} +....+ ^{n}C_{m}
[∵ ^{n}C_{r + 1} + ^{n}C_{r}=^{ n+1}C_{r + 1}]
= ^{m + 3}C_{m + 1} + ^{m + 3}C_{m} +....+ ^{n}C_{m}
Combining in the same way we get
= ^{n}C_{m + 1 }+ ^{n}C_{m }= ^{n + 1}C_{m + 1}= R.H.S.
Again we have to prove ^{n}C_{m + 2} ^{n – 1}C_{m +3 }^{n – 2}C_{m }+....+ (n – m + 1) ^{m}C_{m} = ^{n + 2}C_{m + 2 }
= [^{n}C_{m} + ^{n – 1}C_{m + }^{n – 2}C_{m} +....+ ^{m}C_{m}] + [ ^{n – 1}C_{m} + ^{n – 2}C_{m} +....+ ^{m}C_{m} ] + [^{ n – 2}C_{m} +....+ ^{m}C_{m} ] +....+ [ ^{m}C_{m}]
[n – m + 1 bracketed terms] = ^{n + 1}C_{m + 1} + ^{n}C_{m + 1 }^{n – 1}C_{m+1 }....+ ^{m + 1}C_{m + 1 }
[using previous result.]
= ^{n + 2}C_{m + 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)] < 4n^{2} + 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 = r^{2}
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 b^{2}  4ac > 0 and let α_{1} = c. Prove by induction that
is well  defined and
(Here, ‘well  defined’ means that the denominator in the expression for α_{n + 1} is not zero.) (2001  5 Marks)
Ans. Sol. We have a, b, c the +ve real number s.t. b^{2} – 4ac > 0; α_{1}= c.
is well defined and
For n = 1,
Now, b^{2} – 4ac > 0 ⇒ b^{2 }– 2ac > 2ac > 0
∴ α^{2 }is 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,
∴α_{k + 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)^{2}λ 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 p_{n }denote the probability that no two (or more) consecutive heads occur. Prove that p_{1}=1, p_{2}=1–p^{2 }and p_{n}=(1– p). p_{n–1 }+ p(1 – p) p_{n–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
x^{2}– (1 – p) x–p (1– p)=0 and
Ans. Sol. We have α + β = 1 – p and αβ = – p (1– p)
For n = 1, p_{n} = p_{1 }= 1
Also,
For n = 2, p_{2} = 1– p^{2}
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 .
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