a) \( 1+2+3+4+5+6 \) b) \( 1+4+9+16+25+36+49 \) c) \( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots+\frac{1}{10} \) d) \( \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\ldots+\frac{20}{21} \)

To solve \( 1+2+3+4+5+6 \), simply add the numbers together. This gives you \( 21 \). You could also use the formula for the sum of the first \( n \) natural numbers, which is \( \frac{n(n+1)}{2} \). Here, \( n = 6 \), so \( \frac{6(6+1)}{2} = 21 \). For \( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots+\frac{1}{10} \), you’re dealing with a harmonic series up to the 10th term. While there’s no simple formula for the sum of a harmonic series, you can calculate it step by step. The sum turns out to be approximately \( 2.928968 \). It’s a fun way to see how fractions can sneak up on you!