## Chapter 4 Principle of Mathematical Induction Lecture 3

Question 19. $n (n + 1) (n + 5)$ is a multiple of 3.

Question 20. $10^{2n-1}+1$ is divisible by 11.

Question 21. $x^{2n}-y^{2n}$ is divisible by $x+y$.

Question 22. $3^{2n+2}-8n-9$ is divisible by 8.

Question 23. $41^n-14^n$ is a multiple of 27.

Question 16. $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+…+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$

Question 17. $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+…+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

Question 18. $1+2+3+…+n<\frac{1}{8}(2n+1)^2$

Question 24. $(2n+7)<(n+3)^2$

Example 2. $2^n>n$

Example 7. $1^2+2^2+…+n^2>\frac{n^3}{3}, n \in N$

## Mathematical Induction Lecture 2 Chapter 4

Question 6. $1.2+2.3+3.4+…+n.(n+1)= \left [ \frac{n(n+1)(n+2)}{3} \right ]$

Question 7. $1.3+3.5+5.7+…+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$

Question 8. $1.2+2.2^2+3.2^3+…+n.2^n=(n-1)2^{n+1}+2$

Question 9. $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+…+\frac{1}{2^n}=1-\frac{1}{2^n}$

Question 10. $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+…+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

Question 11. $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+…+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

Question 12. $a+ar+ar^2+…+ar^{n-1}=\frac{a(r^n-1)}{r-1}$

Question 13. $\left ( 1+ \frac{3}{1} \right ) \left ( 1+ \frac{5}{4} \right ) \left ( 1+ \frac{7}{9} \right ) … \left ( 1+ \frac{(2n+1)}{n^2} \right ) = (n+1)^2$

Question 14. $\left ( 1+ \frac{1}{1} \right ) \left ( 1+ \frac{1}{2} \right ) \left ( 1+ \frac{1}{3} \right ) … \left ( 1+ \frac{1}{n} \right ) = (n+1)$

Question 15. $1^2+3^2+5^2+…+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}$

## Class 11 Maths PMI Lecture 1

Introduction to Principle of mathematical induction, explanation using dominos

Proof of Arithmetic Progression using PMI
$a+(a+d)+(a+2d)+…+(a+(n-1)d) = \frac{n}{2} (2a+(n-1)d)$

Question 1. $1+3+3^2+…+3^{n-1} = \frac{(3^n-1)}{2}$

Question 2. $1^3+2^3+3^3+…+n^3= \left ( \frac{n(n+1)}{2} \right )^2$

Question 3. $1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+…+\frac{1}{(1+2+3+…+n)}=\frac{2n}{(n+1)}$

Question 4. $1.2.3+2.3.4+…+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

Question 5. $1.3+2.3^2+3.3^3+…+n.3^n=\frac{(2n-1)3^{n+1}+3}{4}$