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Mathematical Induction — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsMathematical Induction5 Marks
Question
Prove that $\frac{\text{n}^2}{7}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{\text{n}^2}{2}-\frac{37}{210}$ n is a positive integer for all $\text{n}\in\text{N}.$
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Answer

Let p(n): $\frac{\text{n}^2}{7}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{\text{n}^2}{2}-\frac{37}{210}$ n is a positive integer
For n = 1
$\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}$
$=\frac{30+42+70+105-37}{210}$
$=\frac{247-37}{210}$
It is a positive integer
⇒ p(n) is true for n = 1
Let p(n) is true for n = k,
$\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}$ k is positive integer
$\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}\text{k}=\lambda$
For n = k + 1,
$\frac{(\text{k+1})^7}{7}+\frac{(\text{k+1})^5}{5}+\frac{(\text{k+1})^3}{3}+\frac{(\text{k+1})^2}{2}-\frac{37}{210}(\text{k+1})$
$=\frac{1}{7}\big[\text{k}^7+7\text{k}^6+21\text{k}^5+35\text{k}^3+21\text{k}^2+7\text{k}+1\big]\\+\frac{1}{5}\big[\text{k}^5+5\text{k}^4+10\text{k}^3+10\text{k}^2+21\text{k}^2+5\text{k}+1\big]\\+\frac{1}{3}\big[\text{k}^3+3\text{k}^2+3\text{k}+1\big]+\frac{1}{2}\big[\text{k}^2+2\text{k}+1\big]-\frac{37\text{k}}{210}-\frac{37}{210}$
$=\Big[\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}\Big]+\Big[\text{k}^6+3\text{k}^5+5\text{k}^4+3\text{k}^2+\text{k}\\\ \ \ \ \ \ +\frac{1}{7}+\text{k}^4+2\text{k}^3+2\text{k}^2+\frac{1}{5}+\text{k}^2+\text{k}+\frac{1}{3}+\text{k}+\frac{1}{2}-\frac{37}{210}\Big]$
$=\lambda+\text{k}^6+3\text{k}^5+6\text{k}^4+7\text{k}^3+6\text{k}^2+3\text{k}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}$
$=\lambda+\text{k}^6+3\text{k}^5+6\text{k}^4+7\text{k}^3+6\text{k}^2+3\text{k}+1$
= Positive integer
p(n) is true for n = k + 1
p(n) is true for all by $\text{n}\in\text{N}$ PMI

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