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JEE Main Test Series- Mathematical Induction, Principles, Previous Year Question and Solution
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Mathematical Induction topic is one of the least weighing topics in JEE Main Mathematics syllabus. The topic of Mathematical Induction covers approximately 4-8 marks of the entire paper with 1 or 2 questions on the topic which involve proving statements and equations. It is one of the topics that are covered in JEE Main however are not included in JEE Advanced exam. 

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Mathematical Induction is a mathematical technique and a way of proving a statement or formula or any theorem. It is a technique, generally used for proving results or establishing statements for natural numbers. 

Students can use the previous year question papers to prepare for JEE Main Mathematical Induction topic. For reference, a key of previous year questions and solutions are given below for the topic of Mathematical Induction which are more frequently asked in JEE Main Question Paper

Beginners are suggested to use authentic study materials for the same like the NCERTs of mathematics and textbooks of class 10 – 12. After that they could definitely reach out to the below-given set of questions for more thorough revision. 

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JEE Main Test Series Mathematical Induction: Previous Year Questions and Solution

Majorly all questions carry 4 marks and are of year 2016, 2017, 2018 and 2019.

Q.1: If (1 + ax)n = 1 + 8x + 24x2 +…., then the value of a and n are ________.

Solution:
As given (1 + ax)n = 1 + 8x + 24x2 +….
1 + \frac{n}{1}
1
n
ax + \frac{n(n-1)}{1.2}
1.2
n(n−1)
a2x2 + . . . . = 1 + 8x + 24x2 + . . . .
On comparing same coefficients, we have
⇒ na = 8, \frac{n(n-1)}{1.2}
1.2
n(n−1)
a2 = 24
⇒ na (n − 1) a = 48
8 (8 − a) = 48
8 − a = 6
⇒ a = 2
⇒ n = 4

Q.2: If the sum of the coefficients in the expansion of (1 − 3x + 10x2)n is a and if the sum of the coefficients in the expansion of (1 +x2)n is b, then what is the relation between a and b?

Solution:
We have a = sum of the coefficient in the expansion of (1 − 3x + 10x2)n = (1 − 3 + 10)n = (8)n
(1 − 3x + 10x2)n = (2)3n, [Putting x = 1] .
Now, b = sum of the coefficients in the expansion of (1 + x2)n = (1 + 1)n = 2n.
Clearly, a = b3

Q.3: In the polynomial (x − 1) (x − 2) (x − 3) . . . . . (x − 100), what is the coefficient of x99?

Solution:
(x − 1) (x − 2) (x − 3) . . . . . (x − 100)
Number of terms = 100
Coefficient of x99 = (x − 1) (x − 2) (x − 3) . . . . . (x − 100)
= (−1 −2 −3 −. . . . . . −100)
= − (1 + 2 + . . . . . +100)
= − [100 * 101 / 2
= – 5050

Q.4: If the coefficient of x7 in (ax2 + [1/bx])11 is equal to the coefficient of x−7 in (ax − 1/bx2)11, then ab = __________.

Solution:
In the expansion of (ax2 + [1/bx])11, the general term is
Tr + 1 = 11Cr (ax2)11 − r (1/bx)r
= 11Cr * a11 − r * [ 1/br ] * x22 − 3r
For x7, we must have 22 – 3r = 7
r = 5, and the coefficient of x7 = 11C5 * a11−5 * [1/b5] = 11C5 * a6 * b5
Similarly, in the expansion of (ax−1bx2)11, the general term is
Tr + 1 = 11Cr * (−1)r * [a11 − r/br ] * x11 − 3r
For x-7 we must have, 11 – 3r = -7 , r = 6, and the coefficient of x−7 is 11C6 * [a5/b6] = 11C5 * a5 * b6.
As given, 11C5 [a6/b5] = 11C5 * [a5/b6]
⇒ ab = 1

Q.5: In the expansion of (x + a)n, the sum of odd terms is P and sum of even terms is Q, then the value of (P2 − Q2) will be _________.

Solution:
(x + a)n = xn + nC1xn−1 a + . . . . . = (xn + nC2xn−2a2 + . . . . . . . + (nC1xn−1a + nC3xn−3a3 + . . . . .) = P+Q
(x − a)n = P − Q
As the terms are altered,
P2 − Q2 = (P + Q) (P − Q) = (x + a)n (x − a)n
P2 − Q2 = (x2 − a2)n

Q.6: The last digit in 7300 is __________.

Solution:
We have 72 = 49 = 50 − 1
Now, 7300 = (72)150
= (50 − 1)150
= 150C0 (50)150 (−1)0 + 150C1 (50)149 (−1)1 + . . . . . .+ 150C150 (50)0 (−1)150
Thus, the last digits of 7300 are 150C150 * 1 * 1 i.e., 1.

Q.7: The digit in the unit place of the number (183!) + 3183 is _____.

Solution:
We know that n! terminates in 0 for n³ at 5 and 34n terminator in 1, (because 34 = 81)
Therefore, 3180 = (34)45 terminates in 1.
Also 33 = 27 terminates in 7
3183 = 3180 * 33 terminates in 7.
183! + 3183 terminates in 7 i.e. the digit in the unit place = 7

Q.8: The interval in which x must lie so that the greatest term in the expansion of (1 + x)2n has the greatest coefficient, is _____.

Solution:
Here the greatest coefficient is 2nCn.
Therefore, 2nCnxn > 2nCn+1xn−1 ⇒ x > n / [n + 1] and 2nCnxn > 2nCn-1xn+1
⇒ x < [n + 1] / n
Hence, the required interval is (n / [n + 1], [n + 1] / n).

Q.9: Let R = (5√5 + 11)2n + 1 and f = R − [R], where [.] denotes the greatest integer function. The value of R * f is

Solution:
Since (5√5 − 11) (5√5 + 11) = 4
5√5 − 11 = 4 / (5√5 + 11), Because 0 < 5√5 − 11 < 1 ⇒ 0 < (5√5 − 11)2n + 1 < 1, for positive integral n.
Again, (5√5 + 11)2n + 1 − (5√5 − 11)2n + 1 = 2 {2n+1C1 (5√5)2n * 11 + 2n + 1C3(5√5)2n − 2 × 113 + . . .. +
2n + 1C2n+1 112n+1}
= 2 {2n+1C1(125)n * 11 + 2n+1C3(125)n−1 113 +. . . . . . + 2n+1C2n+1 112n+1}
= 2k, (for some positive integer k)
Let f′= (5√5 − 11)2n + 1 , then [R] + f − f′=2k
f − f′ = 2k − [R]
⇒ f − f′ is an integer.
But, 0 ≤ f < 1; 0 < f′<1
⇒ −1 < f − f′ < 1
f − f′ = 0 (integer)
f = f′
Therefore, Rf = Rf′ = (5√5 + 11)2n + 1 * (5√5 − 11)2n + 1
= ([5√5]2 + 112)2n + 1
= 42n+1

Q.10: (1 + x)n − nx − 1 is divisible by (where n∈N)

  1. 2x
  2. x2
  3. 2x3
  4. All of these

Solution:

(1 + x)n = 1 + nx + ([n (n − 1)] / [2!]) * x2 + ([n (n − 1) (n − 2)] / [3!]) * x3 + . . . . .
(1 + x)n − nx − 1 = x2 [([n (n − 1)] / [2!]) + ([n (n − 1) (n − 3)] / [3!]) * x + . . . . .]
From above it is clear that (1 + x)n − nx − 1 is divisible by x2.
Trick: (1 + x)n − nx − 1, put n = 2 and x = 3;
Then 42 − 2 * 3 − 1 = 9 is not divisible by 6, 54 but divisible by 9, which is given by option (b) i.e., x2 = 9

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Mathematical Induction: Important Principles

Below mentioned are some of the important principles based on JEE Main Mathematical Induction which are necessary for the robust and effective preparation of the topic in JEE Main examination. Mathematical induction is used to prove that the given statement is true or not. It uses 2 steps to prove it.

First Principle of Mathematical Induction

  • Base Case: The given statement is correct for first natural number that is, for n=1, p (1) is true.
  • Inductive Step: If the given statement is true for any natural number like n = k then it will be correct for n = k + 1 also that is, if p (k) is true then p (k+1) will also be true.

The first principle of mathematical induction says that if both the above steps are proven then p (n) is true for all natural numbers.

Second Principle of Mathematical Induction

It is more powerful than the first principle. It is sometimes called as strong induction. It doesn’t need a base case in special case otherwise it may require to show with more than one base case. In the recursive step to prove that k +1 is true, first we need to prove that the statement is correct for all the numbers less than k+1 also.

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