Answer to Question #224713 in Algebra for Sneha

Question

Continue the two sequence of numbers below and find an equation to each of the sequence
N 1,2,3,4,5,6,7
an 2,5,9,14,20,27,___
bn 1,3,12,60,360,2520,___
Find equation

Answer 1

The sequences are given as ,

a_n=2,5,9,14,20,27 .....b_n=1,3,12,60,360,2530 ........ 

First sequence is

a_n=2,5,9,14,20,27 .....

So ,

a₁=2 , a₂=5 , a₃=9 , a₄=14

Now ,

a₂−a₁=3

a₃−a₂=4

a₄−a₃=5

·········

a_n−a_n-1=n+1

This is cumulative sequence. Therefore by adding we get ,

a_n−a₁=3+4+5+·······+(n+1)

⇒a_n−2=3+4+5+·······+(n+1)

⇒a_n=2+3+4+5+·······+(n+1)

⇒a_n=$\frac{n}{2}[2+(n+1)]$

⇒a_n=$\frac{n(n+3)}{2}$

put n=7 ,

a₇=$\frac{7(7+3)}{2}=\frac{7×10}{2}=35$

Thus , the equation is a_n=$\frac{n(n+3)}{2}$

and a₇=35

And second sequence is

b_n=1,3,12,60,360,2530 ........

So ,

b₁=1 , b₂=3 , b₃=12 , b₄=60

Now ,

$\frac{b_2}{b_1}$ =3

$\frac{b_3}{b_2}=4$

$\frac{b_4}{b_3}=5$

······

$\frac{b_n}{b_{(n-1)}}=n+1$

This is multiplicative sequence , Therefore by multiply we get ,

$\frac{b_2}{b_1}×\frac{b_3}{b_2}×\frac{b_4}{b_3}×...........\frac{b_n}{b_{n-1}}=3×4×5.......×(n+1)$

$\frac{b_n}{b_1}=3×4×5×.....×(n+1)$

$\implies\frac{b_n}{1}=3×4×5×.........×(n+1)$

$\implies b_n=3×4×5×.........×(n+1)$

b_n=$\frac{1×2×3×4×5×......×(n+1)}{2}$

$\implies b_n=\frac{(n+1)!}{2}$

put n=7,

b₇=$\frac{(7+1)!}{2}=\frac{8!}{2}=\frac{40320}{2}=20160$

Thus , the equation is b_n=$\frac{(n+1)!}{2}$ and b₇=20160