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Mathematical Induction — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsMathematical Induction5 Marks
Question
Prove that $\stackrel{{7+77+777+...+777\ ...................\ 7}}{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox{ n-digits}}$ $$${ =\frac{7}{81}(10^{\text{n}+1}-9\text{n}-10)}$ for all $\text{n}\in\text{N}.$
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Answer

Let p(n) be the statement given by
p(n): $7+77+777+...+777\ ........\ 7 { =\frac{7}{81}(10^{\text{n}+1}-9\text{n}-10)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{n-digits}$ for all $\text{n}\in\text{N}.$
Step 1:
P(1) : $7-\frac{7}{81}[10^{1+1}-9(1)=10]$
$\Rightarrow7=\frac{7}{81}\times(100-9-10)$
$\Rightarrow7=\frac{7}{81}\times81$
$\Rightarrow7=7\times(1)$
$\therefore$ P(1) is true.
Step 2:
Let p(m) is true. Then,
$7+77+777+...+777\ ........\ 7 { =\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m-digits}$
We have to prove that p(m + 1) is true.
$7+77+777+...+777\ ........\ 7=7+77+777+...+777\ ........\ 7+777..............7 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m+1-digits} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m-digits}\ \ \ \ \ \ \ \ \ \ \ \ \text{m+1-digits}$
${ =\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)}+{7{[1111...............1]}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\text{m+1-digits}}$
${ =\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)}+{7{[1111...............1]}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m+1-digits}$
${ =\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)}+{7{[9999...............9]}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m+1-digits}$
$=\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)+\frac{7}{9}[10^{\text{m}+1}-1]$
$=\frac{7}{81}\big[(1+9)10^{\text{m+1}}-9\text{m}-19\big]$
$=\frac{7}{81}\big[10\times10^{\text{m}+1}-9(\text{m+1)}-10\big]$
$=\frac{7}{81}\big[10\times10^{\text{m}+2}-9(\text{m+1)}-10\big]$
⇒ p(m + 1) is true.

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