Question

find the sum of all terms in the given sequence 1, 2+x, 3+x^2, 4+x^3....... n+x^n-1​

Answer 1

Answer:

n(n + 1 ) / 2 + x ( xⁿ⁻¹ - 1 ) / ( x - 1 )

Step-by-step explanation:

Given---> Sequence,

1 , (2 + x ) , (3 + x² ) , ...................( n + xⁿ⁻¹ )

To find---> Sum of all terms of given sequence

Solution--->

( 1 ) + ( 2 + x ) + ( 3 + x² )+.............+ ( n + xⁿ⁻¹ )

Rearranging the terms, we get,

= ( 1 + 2 + 3 + ........+ n ) + (x + x² + x³ + ........+ xⁿ⁻¹ )

Now we solve both bracket one by one,

First bracket,

1+2+3+ .................+n

Clearly its an AP whose first term is 1 and common difference is also one and number of terms are n

Now formula of sum of n terms of AP is,

Sₙ = n/2 {2a + ( n - 1 ) d }

a = 1 , d = 1 , n = n

=> Sₙ = ( n/2 ) {2 ( 1 ) + (n - 1 ) ( 1 ) }

= ( n/2 ) { 2 + n - 1 )

= ( n/2 ) ( n + 1 )

= n ( n + 1 ) / 2

Now , we solve second bracket,

x + x² + x³ + ..................+ xⁿ⁻¹

Clearly it is a GP , whose first term is x and common ratio is also x and it consist of ( n - 1 ) terms

Formula of sum of n terms of a GP is,

S¹ₙ = a ( rⁿ - 1 ) / ( r - 1 )

Here , a = x , r = x , n =( n - 1 )

S¹ₙ = x ( xⁿ⁻¹ - 1 ) / ( x - 1 )

Now sum of given sequence is

= Sₙ + Sₙ¹

= { n ( n + 1 ) / 2 } + x { ( xⁿ⁻¹ - 1 ) / ( x - 1 ) }

#Answerwithquality

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