Question

prove it by mathematical induction​

Answer 1

Step-by-step explanation:

Given :

To prove that,

$7 + 77 + 777 + ... 7..(n \ terms) = \frac{7}{81} (10^{n+1} - 9n - 10)$

By mathematical induction.

Proof :

let  $P(n) =7 + 77 + 777 + ... 7..(n \ terms) = \frac{7}{81} (10^{n+1} - 9n - 10)$

If n = 1,

$P(1) = 7= \frac{7}{81} (10^{1+1} - 9(1) - 10)}$

$P(1) = 7= \frac{7}{81} (100 - 9 - 10)}$

$P(1) = 7= \frac{7}{81} (81)}$

$P(1) = 7= 7$

⇒ LHS = RHS,

Hence, P(n) is true for n = 1,

Let P(n) be n = k,

Then,

$7 + 77 + 777 + ... 7..(k \ terms) = \frac{7}{81} (10^{k+1} - 9k - 10)$

be true,,

To Prove :

P(n) for n  = k + 1,

i.e.,

$7 + 77 + 777 + ... 7..(k+1 \ terms) = \frac{7}{81} (10^{(k+1) + 1} - 9(k+1) - 10)$

Proof :

LHS = $[7 + 77 + 777 + ... 7..(k \ terms)] + 7..(k+1 \ terms)$

= $\frac{7}{81} (10^{k+1} - 9k - 10) + 7..(k+1 \ terms)$

= $\frac{7}{81} (10^{k+1} - 9k - 10) +(\frac{7}{9}) 9..(k+1 \ terms)$

= $\frac{7}{81} (10^{k+1} - 9k - 10) +(\frac{7}{9}) (10 - 1)..(k+1 \ terms)$

= $\frac{7}{81} (10^{k+1} - 9k - 10) +(\frac{7}{9})(10^{k+1} - 1)$

= $\frac{7}{81} (10^{k+1} - 9k - 10) +(\frac{7}{9 \times 9})(9)[10^{k+1} - 1]$

= $\frac{7}{81} (10^{k+1} - 9k - 10) +(\frac{7}{81})[9 \times 10^{k+1} - 9]$

= $\frac{7}{81} (10^{k+1} - 9k - 10 + [9 \times 10^{k+1} - 9])$

= $\frac{7}{81} ([9 +1]10^{k+1} - 9(k + 1) - 10)$

= $\frac{7}{81} ([10]10^{k+1} - 9(k + 1) - 10)$

= $\frac{7}{81} (10^{k+2} - 9(k + 1) - 10)$

= $\frac{7}{81} (10^{(k+1) + 1} - 9(k + 1) - 10)$ = RHS

Hence, proved.