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