Inductive Step
Trending Questions
Q.
Let be a function defined by , where denotes the greatest integer . Then the range of is
Q. If the sum of an infinite GP a, ar, ar2, ar3, ... is 15 and the sum of the squares of its each term is 150, then the sum of ar2, ar4, ar6, ... is
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Q. The positive integral values of n such that 1.21+2.22+3.23+4.24+5.25+.....+n.2n=2(n+10)+2 is
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Q.
Let be a finite set containing elements. Then, the total number of commutative binary operation on is:
Q. Prove the following statement by using the principle of mathematical induction for all n∈N
a+ar+ar2+⋯+arn−1=a(rn−1)r−1
a+ar+ar2+⋯+arn−1=a(rn−1)r−1
Q. Prove the following by using the principle of mathematical induction for all n∈N
1+1(1+2)+1(1+2+3)+⋯+1(1+2+3+⋯+n)=2nn+1
1+1(1+2)+1(1+2+3)+⋯+1(1+2+3+⋯+n)=2nn+1
Q.
Evaluate the expression .
Q. Prove the following by using the principle of mathematical induction for all n∈N.
13+23+33+⋯+n3=(n(n+1)2)2
13+23+33+⋯+n3=(n(n+1)2)2
Q. Prove the following by using the principle of mathematical induction for all n∈N
12⋅5+15⋅8+18⋅11+⋯+1(3n−1)(3n+2)=n(6n+4)
12⋅5+15⋅8+18⋅11+⋯+1(3n−1)(3n+2)=n(6n+4)
Q. Prove the following by using the principle of mathematical induction for all n∈N.
1+3+32+⋯+3n−1=(3n−1)2.
1+3+32+⋯+3n−1=(3n−1)2.
Q. Prove the following by using the principle of mathematical induction for all n∈N.
13⋅5+15⋅7+17⋅9+⋯+1(2n+1)(2n+3)=n3(2n+3)
13⋅5+15⋅7+17⋅9+⋯+1(2n+1)(2n+3)=n3(2n+3)
Q. Prove the following by using the principle of mathematical induction for all n∈N
1⋅3+2⋅32+3⋅33+⋯+n⋅3n=(2n−1)3n+1+34
1⋅3+2⋅32+3⋅33+⋯+n⋅3n=(2n−1)3n+1+34
Q. Prove the following by using the principle of mathematical induction for all n∈N
1⋅3+3⋅5+5⋅7+⋯+(2n−1)(2n+1)
=n(4n2+6n−1)3
1⋅3+3⋅5+5⋅7+⋯+(2n−1)(2n+1)
=n(4n2+6n−1)3
Q.
Evaluate the following.
for
Q. Prove the following by using the principle of mathematical induction for all n∈N.
12+32+52+⋯+(2n−1)2=n(2n−1)(2n+1)3
12+32+52+⋯+(2n−1)2=n(2n−1)(2n+1)3
Q. Prove the following by using the principle of mathematical induction for all n∈N.
1+2+3+4+⋯+n<18(2n+1)2.
1+2+3+4+⋯+n<18(2n+1)2.
Q. Let A2×m and Bn×2 are two matrices such that AB and BA exist. If C is a matrix such that the order of matrix BAC is 3×4, then which of the following options is/are true
- m=2, n=3
- m=n=3
- order of matrix C must be 3×4
- matrix addition is not possible between the matrices A, B and C
Q. Prove the following by using the principle of mathematical induction for all n∈N
12+14+18+⋯+12n=1−12n
12+14+18+⋯+12n=1−12n
Q. Prove the following by using the principle of mathematical induction for all n∈N
(1+31)(1+54)(1+79)⋯(1+(2n+1)n2)=(n+1)2
(1+31)(1+54)(1+79)⋯(1+(2n+1)n2)=(n+1)2
Q. Prove the following by using the principle of mathematical induction for all n∈N.
11⋅4+14⋅7+17⋅10+⋯+1(3n−2)(3n+1)=n(3n+1)
11⋅4+14⋅7+17⋅10+⋯+1(3n−2)(3n+1)=n(3n+1)
Q. Prove the following by using the principle of mathematical induction for all n∈N
11⋅2⋅3+12⋅3⋅4+13⋅4⋅5+⋯+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)
11⋅2⋅3+12⋅3⋅4+13⋅4⋅5+⋯+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)
Q. For every positive integral value of n, 3n>n3 when
Q. 12+22+…+n2>n33, n∈N
Q. If a1, a2, a3, ..., an are in A.P. with sn as the sum of first 'n' terms (S0=0), then ∑nk=0nCkSk is equal to
Q. For all positive integral values of n, 32n−2n+1 is divisible by
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Q. Using principle of Mathematical induction prove that:
x2n−y2n is divisible by x+y, where n∈N
x2n−y2n is divisible by x+y, where n∈N
Q.
Let P(n) : 2n<(1×2×3×....×n). Then the smallest positive integer for which P(n) is true is
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Q. Prove the following by using principle of mathematical induction for all n∈N :
1+1(1+2)+1(1+2+3)+.......+1(1+2+3+…n)=2n(n+1).
1+1(1+2)+1(1+2+3)+.......+1(1+2+3+…n)=2n(n+1).
Q. Let P(n)=n3−n, the largest number by which P(n) is divisible ∀ possible integral values of n is
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Q. Using the principle of mathematical induction, prove the following for all n∈N:
12+32+52+72+....+(2n−1)2=n(2n−1)(2n+1)3
12+32+52+72+....+(2n−1)2=n(2n−1)(2n+1)3