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Question
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)
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Solution
Step (1): Assume given statement
Let the given statement be P(n), i.e.,
P(n)=11⋅4+14⋅7+17⋅10+⋯+1(3n−2)(3n+1)=n(3n+1)
Step (2): Checking statement P(n) for n=1
Put n=1 in P(n), we get
P(1):11⋅4=13⋅1+1
⇒14=14
Thus P(n) is true for n=1
Step (3): P(n) for n=K.
Put n=K in P(n) and assume this is true for some natural number K i.e.,
P(k):11⋅4+14⋅7+17⋅10+⋯+1(3K−2)(3K+1)=K(3K+1) ⋯(1)
Step (4): Checking statement P(n) for n=K+1
Now, we shall prove that P(K+1) is true whenever P(K) is true.
Now we have
11⋅4+14⋅7+17⋅10+⋯+1(3K−2)(3K+1)+1{3(K+1)−2}{3(K+1)+1}
=K(3K+1)+1(3K+1)(3K+4) (using (1))
=1(3K+1){K+1(3K+4)}
=1(3K+1){3K2+4K+13K+4}
=1(3K+1){3K2+3K+K+13K+4}
=1(3K+1){(K+1)(3K+1)3K+4}
=K+13K+4
We can write it as
(K+1){3(K+1)+1}
Thus, P(K+1) is true whenever P(K) is true.
Final answer:
Therefore, by the principle of mathematical induction, statement P(n) is true for all n∈N.
Let the given statement be P(n), i.e.,
P(n)=11⋅4+14⋅7+17⋅10+⋯+1(3n−2)(3n+1)=n(3n+1)
Step (2): Checking statement P(n) for n=1
Put n=1 in P(n), we get
P(1):11⋅4=13⋅1+1
⇒14=14
Thus P(n) is true for n=1
Step (3): P(n) for n=K.
Put n=K in P(n) and assume this is true for some natural number K i.e.,
P(k):11⋅4+14⋅7+17⋅10+⋯+1(3K−2)(3K+1)=K(3K+1) ⋯(1)
Step (4): Checking statement P(n) for n=K+1
Now, we shall prove that P(K+1) is true whenever P(K) is true.
Now we have
11⋅4+14⋅7+17⋅10+⋯+1(3K−2)(3K+1)+1{3(K+1)−2}{3(K+1)+1}
=K(3K+1)+1(3K+1)(3K+4) (using (1))
=1(3K+1){K+1(3K+4)}
=1(3K+1){3K2+4K+13K+4}
=1(3K+1){3K2+3K+K+13K+4}
=1(3K+1){(K+1)(3K+1)3K+4}
=K+13K+4
We can write it as
(K+1){3(K+1)+1}
Thus, P(K+1) is true whenever P(K) is true.
Final answer:
Therefore, by the principle of mathematical induction, statement P(n) is true for all n∈N.
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