Question

find the least positive integer n such that 1+6+6²+.......+6n>5000​

Answer 1

Answer:

9ki power 4

Step-by-step explanation:

ok now I will write answers

Answer 2

Concept:

Geometric Progression (GP): It is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio.

$S_{n}=\frac{a(r^{n}-1)}{r-1}$

where,

$S_{n}=Sum\ of\ nth\ terms\ in\ GP$

r is common ratio

a is the first time

Given:

The below is given

1+6+6²+.......+6n>5000​

To find:

We need to find the least positive integer n

Solution:

Now, using GP

1+6+6²+.......+6n

$r=\frac{6}{1} =6$ >1

Now since r > 1

a = 1

$S_{n}=\frac{a(r^{n}-1)}{r-1}=\frac{1(6^{n}-1)}{6-1}$

⇒$S_{n}=\frac{6^{n}-1}{5}$            ------------(1)

Now, from equation (1)

$S_{n}=\frac{6^{n}-1}{5} > 5000$

⇒$\frac{6^{n}-1}{5}*5 > 5000*5$

⇒${6^{n}-1} > 25000$

⇒$6^{n} > 25000+1$

⇒$6^{n} > 25001$

Now by substituting values of n=1, 2, 3, 4, 5, 6, 7.....

When

$n=1, 6^{1} > 25001$                                          [false]

$n=2,\ 6^{2} > 25001=36 > 25001\ \ \ \ \ \ \ \ \ \ [false]\\n=3,\ 6^{3} > 25001=216 > 25001\ \ \ \ \ \ \ \ [false]\\ n=4,\ 6^{4} > 25001=1296 > 25001\ \ \ \ \ \ \ \ \ [false]\\n=5,\ 6^{5} > 25001=7776 > 25001\ \ \ \ \ \ \ \ \ [false]\\n=6,\ 6^{6} > 25001=46656 > 25001 \ \ \ \ \ \ \ \ \ [true]$

Therefore the least positive integer n is 6