+0

# Pattern

+1
240
2
+597

Hi guys, me again,

last one for the day...

T1=2, T2=6 and T3=12. Determine the nth term formula.

so I said:

T1=2

T2=6

T3=12

But:

2=1.2

6=2.3

12=3.4

This gives me the formula: Tn = n(n+1)

My problem is, they ask to determine which term will have a value of 420?....I cannot use the formula I got to...so I must have gone wrong somewhere. Could someone please show me my error?...thank you all kindly..

Mar 5, 2019

#1
+23856
+3

T1=2, T2=6 and T3=12. Determine the nth term formula.

$$\begin{array}{|rcll|} \hline t_1 = 1 \cdot 2 &=& 2 \\ t_2 = 2 \cdot 3 &=& 6 \\ t_3 = 3 \cdot 4 &=& 12 \\ \ldots \\ t_n &=& n(n+1) \\ \hline 420 &=& n(n+1) \\ 420 &=& n^2+n \\ n^2+n -420 &=& 0 \\ n&=& \dfrac{-1\pm\sqrt{1-4\cdot(-420)}}{2} \\ n&=& \dfrac{-1\pm\sqrt{1681}}{2} \\ n&=& \dfrac{-1\pm 41}{2} \\ n&=& \dfrac{-1 {\color{red}+} 41}{2} \quad | \quad n > 0\ ! \\ \mathbf{n} &\mathbf{=}& \mathbf{ 20 }\\ \hline \end{array}$$

$$\mathbf{t_{20} = 20\cdot 21 = 420}$$

Mar 5, 2019

#1
+23856
+3

T1=2, T2=6 and T3=12. Determine the nth term formula.

$$\begin{array}{|rcll|} \hline t_1 = 1 \cdot 2 &=& 2 \\ t_2 = 2 \cdot 3 &=& 6 \\ t_3 = 3 \cdot 4 &=& 12 \\ \ldots \\ t_n &=& n(n+1) \\ \hline 420 &=& n(n+1) \\ 420 &=& n^2+n \\ n^2+n -420 &=& 0 \\ n&=& \dfrac{-1\pm\sqrt{1-4\cdot(-420)}}{2} \\ n&=& \dfrac{-1\pm\sqrt{1681}}{2} \\ n&=& \dfrac{-1\pm 41}{2} \\ n&=& \dfrac{-1 {\color{red}+} 41}{2} \quad | \quad n > 0\ ! \\ \mathbf{n} &\mathbf{=}& \mathbf{ 20 }\\ \hline \end{array}$$

$$\mathbf{t_{20} = 20\cdot 21 = 420}$$

heureka Mar 5, 2019
#2
+597
+3

Myyyyyyyyyyyyy ....... gooooooodnesss!!!!!!!..Heureka,

so my formula was indeed correct!!..thank you sooo very much!!..you are one appreciated girl!!!..

juriemagic  Mar 5, 2019