Question

find the missing number in the series 1 , 43 , 161 , 407 ,​

Answer 1

The missing number (?) = 521

Step-by-step explanation:

The given series are 1 , 43 , 161 , 407 ,​ ___?__

There are two series  1 ,  161 , ​ ___?__

and 43 , 407 ,​

The first pattern is ($3^{2}$ - 8) , ($13^{2}$ - 8) , ($23^{2}$ - 8) .

The second pattern is ($6^{2}$ + 6) , ($20^{2}$ + 6).

∴ Missing number (?) = ($23^{2}$ - 8) = 529 - 8

= 521

Hence, missing number (?) = 521

Answer 2

The missing one is $521$

Step-by-step explanation:

Given,

Find the missing number in the series $1 , 43 , 161 , 407 ,?$

In above expression given two different series,

1st one is,

$1,161,?$

Where,

$3^{2}-8=1,13^{2}-8=161$

Similarly, $23^{2}-8=521$

And 2nd one is,

$43,407,?$

Where,

$6^{2}+7=43,20^{2}+7=407$

Similarly, $34^{2}+7=1148$

∴ The series $1 , 43 , 161 , 407 ,521$

So, The missing one is $521$