Question

What is the value of this ?>>>>

Answer 1

Answer:

option D is correct.

Step-by-step explanation:

We need to find the value of

$\sum_{n=1}^{6} 4(3)^{n-1}$

Here value of n starts from 1 and goes on till 6

And we need to add the values of all the terms by putting value of n from 1 to 6

This can be written as:

$=4(3)^{1-1}+4(3)^{2-1}+4(3)^{3-1}+4(3)^{4-1}+4(3)^{5-1}+4(3)^{6-1} \\    Solving\\=4(3)^0+4(3)^1+4(3)^2+4(3)^3+4(3)^4+4(3)^5\\=4(1)+4(3)+4(9)+4(27)+4(81)+4(243)\\=4+12+36+108+324+972\\=1456$

So, option D is correct.

Answer 2

Answer:

1456

Step-by-step explanation:

This is the sum of a geometric sequence

The n th term of a geometric sequence is

$a_{n}$ = a$(r)^{n-1}$

where a is the first term and r the common ratio

4$(3)^{n-1}$ ← is in this form

with a = 4 and r = 3

The sum to n terms of a geometric sequence is

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

$S_{6}$ = $\frac{4(3^6-1)}{3-1}$ = $\frac{4(729-1)}{2}$ = 2 × 728 = 1456