Question

Find the sum of the following series upto n terms 1+5+11+19+....

Answer 1

Step-by-step explanation:

Let the sum of given series be,

Sₙ = 1 + 5 + 11 + 19 + 29 ............tₙ

Sₙ= 0 + 1 + 5 + 11 + 19 + 29 ........tₙ₋₁ + tₙ

Subtracting the both terms according as their corresponding terms,

We get,

0 = 1 + 4 + 6 + 8 + ............ + (tₙ - tₙ₋₁) - tₙ

tₙ = 1 + 4 + 6 + 8 + ..................+ (tₙ - tₙ₋₁)

tₙ = 1 + [4 + 6 + 8 + ..........+(n-1) terms ]

Now consider the sequence,

4 , 6 , 8 , ......................(n-1) terms

This is an Arithmetic Progression, where

First term (a). = 4

Common Difference (d) = 2

Number of terms = n - 1

Thus,

$\text{Sum of Terms}=\frac{n-1}{2}(2a+(n-1-1)d)\\\;\\=\frac{n-1}{2}(2×4+(n-2)2)\\\;\\=\frac{n-1}{2}(8+2n-4)\\\;\\=\frac{(n-1)(2n+4)}{2}\\\;\\=(n-1)(n+2)$

Now,

tₙ = 1 + (n - 1)(n + 2)

tₙ = 1 + (n² + 2n - n - 2)

tₙ = 1 + (n² + n - 2)

tₙ = n² + n - 1

We know that,

$S_n=\sum T_n\\\;\\=\sum(n^2+n-1)\\\;\\=\sum n^2+\sum n-\sum1\\\;\\=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} - n$

$=\frac{n(n+1)}{2}(\frac{2n+1}{3}+1)-n\\\;\\=\frac{n(n+1)}{2}(\frac{2n+4}{3})-n\\\;\\=\frac{n(n+1)(n+2)}{3}-n\\\;\\=\frac{n(n+1)(n+2)-3n}{3}\\\;\\=\frac{n^3+3n^2-n}{3}$

Answer 2

Answer:

n cube +3n square-n/3

It is the answer for the question

Hope it helps u