Question

This is worth 50 points if It is not correct I will report you please I need help

Please find both answers with explanations Find the sum of the 1st n terms
1+2+3+4+5+..... And 4+1+1/4+1/16+1/64+....

Answer 1

Step-by-step explanation:

Case I :

1+2+3+4+5+.........

On observation, it's clear that,

It's an AP, Arithmetic Progression

Here, we have,

• First term, a = 1
• Common Difference, d = (2-1) = 1

Now, We know that, Sum of 'n' terms of an AP is given by the formula,

$S_{n} = \dfrac{n}{2} (2a + (n - 1)d)$

On substituting the values, we get,

$= > S_{n} = \frac{n}{2} (2 \times 1 + 1(n - 1)) \\ \\ = > S_{n} = \dfrac{n}{2} (2 + n - 1) \\ \\ = > \bold{S_{n} = \dfrac{n(n + 1)}{2} }$

Case II

4+1+1/4+1/16+1/64+........

Further, we can resolve this as,

4+(1+1/4+1/16+1/64+.......)

Now, on observing the series (1+1/4+1/16+.....)

We come to know that,

It's a GP, Geometric Progression.

Here, we have,

• First term, a = 1
• Common ratio = (1/4)/1 = 1/4

Now, we know that, sum of n terms of a GP when r ≠ 1, is given by the formula,

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

On substituting the values, we get,

$= > S_{n} = \dfrac{1 \times ( {( \frac{1}{4} )}^{n} - 1)}{( \frac{1}{4} - 1)} \\ \\ = > S_{n} = \dfrac{( \frac{1}{4} )^{n} - 1 }{ - \frac{3}{4} } \\ \\ = > S_{n} = \frac{4}{3} (1 - {4}^{ - n} )$

Hence, overall sum for the series will be,

$= > \bold{S_{n} = 4 + \dfrac{4}{3} (1 - {4}^{ - n} )}$

Answer 2

This is what we call an infinite series. A sum can be found if the series is convergent ie. all the terms add up to approach a certain value. This happens when the terms are continuously decreasing so we can expect them to become zero as the number of terms approach infinity.

The general term for this series is 1/4^n-1.

So we can see that for n=infinity, the general term equals 1/infinity which is zero ie. the given series is convergent and we can find a sum of its terms.

S=1+1/4+1/16+….infinity

=1/(1-1/4)

=4/3