Can you use this formula to find S20, S50 and S1000? Class 9

Can you use this formula to find S₂₀, S₅₀ and S₁₀₀₀? Class 9

Question 1.

Can you use this formula to find S₂₀, S₅₀ and S₁₀₀₀? Class 9

Answer:

Yes, we can use S_n = $\frac{n(n+1)}{2}$ to find S₂₀, S₅₀ and S₁₀₀₀.

$\begin{aligned} & S_{20}=\frac{20 \times 21}{2}=210 \\ & S_{50}=\frac{50 \times 51}{2}=1275 \\ & S_{1000}=\frac{1000 \times 1001}{2}=500500\end{aligned}$

This formula can also be used to find the sum of consecutive natural numbers:

25+ 26+ 27 + ..... + 58

= (1 + 2 + 3 + ... + 58) - (1 + 2 + 3 + ... + 24)

= S₅₈ - S₂₄

= $\frac{58 \times 59}{2}-\frac{24 \times 25}{2}$

= 29 × 59 - 12 × 25

= 1711 - 300 = 1411

Question 2.

Let us revisit the sequence t_n of triangular number, 1, 3, 6, 10, 15,... . Note that the n^th term of this sequence is the sum of the first n natural numbers. Thus t_n = $\frac{n(n+1)}{2}$ Class 9

Can you use this to find the 10^th, 17^th and 80^th triangular number?

Answer:

Yes, to find given triangular numbers,

we use the formula t_n = $\frac{n(n+1)}{2}$ where n is the position of the triangular number in the sequence.

10^th triangular number,

$t_{10}=\frac{10 \times 11}{2}=55$

17^th triangular number,

$t_{17}=\frac{17 \times 18}{2}=153$

80^th triangular number,

$t_{80}=\frac{80 \times 81}{2}=3240$