I find that the harder I work, the more luck I seem to have. – Thomas Jefferson

 Exercise 4.1

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}