1-1-4-1-4-7-1-7-10- - -1-3 n-23 n-1-n-3 n-1

Let P(n) : 1/1.4 + 1/4.7 + 1/7.10 + ..... + 1/(3n 2)(3n + 1) = n/3n + 1 Put n = 1 P(1) : 1/1.4 = 1/4 1/4 = 1/4 ⇒ P(n) is true for n = 1 Let P(n) is true for n ...

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Byju's Answer

Standard XII

Mathematics

Proof by mathematical induction

1/1.4+1/4.7+1...

Question

$\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . . + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{3 n + 1}$

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Solution

Let P(n) : $\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . . + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{3 n + 1}$

Put n = 1

P(1) : $\frac{1}{1.4} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

$\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . . . + \frac{1}{(3 k - 2) (3 k + 1)} = \frac{k}{3 k + 1}$ ........(1)

We have to show that,

$\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . + \frac{1}{(3 k - 2) (3 k + 1)} + \frac{1}{(3 k + 1) (3 k + 4)} = \frac{(k + 1)}{(3 k + 4)}$

Now,

${\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . . . + \frac{1}{(3 k - 2) (3 k + 1)}} + \frac{1}{(3 k + 1) (3 k + 4)}$

$= \frac{k}{(3 k + 1)} + \frac{1}{(3 k + 1) (3 k + 4)}$

$= \frac{1}{(3 k + 1)} [\frac{k}{1} + \frac{1}{(3 k + 4)}]$

$= \frac{1}{(3 k + 1)} [\frac{k (3 k + 4) + 1}{3 k + 4}]$

$= \frac{1}{(3 k + 1)} [\frac{3 k^{2} + 4 k + 1}{(3 k + 4)}]$

$= \frac{1}{(3 k + 1)} [\frac{3 k^{2} + 4 k + 1}{3 k + 4}]$

$= \frac{1}{(3 k + 1)} [\frac{3 k^{2} + 3 k + k + 1}{3 k + 4}]$

$= [\frac{3 k (k + 1) + (k + 1)}{(3 k + 1) (3 k + 4)}]$

$= \frac{(k + 1) (3 k + 1)}{(3 k + 1) (3 k + 4)}$

$= \frac{(k + 1)}{(3 k + 4)}$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true all n $ϵ$ N by PMI.

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Similar questions

Prove $\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \dots + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{(3 n + 1)}$

Q. Prove that

\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{(3 n + 1)}

Q. Prove the following by using the principle of mathematical induction for all

n \in N : \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . . + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{(3 n + 1)}

Prove the following by using the principle of mathematical induction for all n ∈ N:

$\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . .$

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11.4+14.7+17.10+.....+1(3n−2)(3n+1)=n3n+1

$\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . . + \frac{1}{(3 n - 2) (3 n + 1)} = \frac{n}{3 n + 1}$