11-2-12-3-13-4-1 nn-1- nn -1

Let P(n) be the given statement. Now, P(n) #160;=11.2+12.3+13.4+...+1n(n+1)=nn+1Step #160;1:P(1) #160;=11.2=12=11+1Hence, #160;P(1) #160;is #160;tru ...

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Standard XII

Mathematics

Proof by mathematical induction

11.2+12.3+13....

Question

$\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{n (n + 1)} = \frac{n}{n + 1}$

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Solution

Let P(n) be the given statement.
Now,
$P (n) = \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{n (n + 1)} = \frac{n}{n + 1} Step 1 : P (1) = \frac{1}{1.2} = \frac{1}{2} = \frac{1}{1 + 1} Hence, P (1) is true . Step 2 : Let P (m) be true . Then, \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{m (m + 1)} = \frac{m}{m + 1} We shall now prove that P (m + 1) is true . i . e ., \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{(m + 1) (m + 2)} = \frac{m + 1}{m + 2} Now, P (m) = \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{m (m + 1)} = \frac{m}{m + 1} \Rightarrow \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{m (m + 1)} + \frac{1}{(m + 1) (m + 2)} = \frac{m}{m + 1} + \frac{1}{(m + 1) (m + 2)} [Adding \frac{1}{(m + 1) (m + 2)} to both sides] \Rightarrow \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{(m + 1) (m + 2)} = \frac{m^{2} + 2 m + 1}{(m + 1) (m + 2)} = \frac{(m + 1)^{2}}{(m + 1) (m + 2)} = \frac{m + 1}{m + 2} Therefore, P (m + 1) is true . B y t h e p rinciple of m athematical i nduction, P (n) i s t r u e for all n \in N .$

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Similar questions

Q. Write the sum to n terms of the series

\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{n (n + 1)}

Q. The sum of the series

11.2−12.3+13.411.2−12.3+13.4 is equal to

Q. The sum of the series

\frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} \dots \dots \dots

up to

\infty

is equal to

Q. If

n

is a positive integer, show that
(1)

n^{n + 1} - n {(n - 1)}^{n + 1} + \frac{n (n - 1)}{2!} {(n - 2)}^{n + 1} - \dots = \frac{1}{2} n (n + 1)!

;
(2)

n^{n} - (n + 1) {(n - 1)}^{n} + \frac{(n + 1) n}{2!} {(n - 2)}^{n} - \dots = 1

;
the series in each case being extended to

n

terms; and
(3)

1^{n} - n 2^{n} + \frac{n (n - 1)}{1 \cdot 2} 3^{n} - \dots = {(- 1)}^{n} n!

;
(4)

{(n + p)}^{n} - n {(n + p - 1)}^{n} + \frac{n (n - 1)}{2!} {(n + p - 2)}^{n} - \dots = n!

;
the series in the last two cases being extended to

n + 1

terms.

lim n \to \infty ((\frac{n}{n})^{n} + (\frac{n - 1}{n})^{n} + . . + ({\frac{1}{n}}^{n}))

equals

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