Question

Find the first six partial sums S1, S2, S3, S4, S5, S6 of the sequence whose nth term is given. 1 2 , 1 22 , 1 23 , 1 24 , . .

Answer 1

the first partial sum
`\mathbf{S_1= (1)/(2)}`

the second partial sum
`\mathbf{S_2= (3)/(4) }`

the third partial sum
`\mathbf{S_3= (7)/(8)}`

the fourth partial sum
`\mathbf{S_4= (15)/(16)}`

the fifth partial sum
`\mathbf{S_5= (31)/(32)}`

the sixth partial sum
`\mathbf{S_6= (63)/(64)}`

Explanation:

The term of the sequence are given as :
`(1)/(2)`,
`(1)/(2^2)`,
`(1)/(2^3)`,
`(1)/(2^4 )` , . . .

The nth term for this sequence is ,
`\mathtt{a_n =( (1)/(2))^n}`

The nth partial sum of the sequence for
`\mathtt{a_1,a_2,a_3.... a_n}` is
`\mathtt{S_n}`

where;

`\mathtt{S_n = a_1 +a_2+a_3+ ...+a_n}`

The first partial sum is:
`\mathtt{S_1= a_1}`

`\mathtt{S_1= ((1)/(2))^1}`

`\mathbf{S_1= (1)/(2)}`

Therefore, the first partial sum
`\mathbf{S_1= (1)/(2)}`

The second partial sum is:
`\mathtt{S_2= a_1+a_2}`

`\mathtt{S_2= ((1)/(2))^1 + ((1)/(2))^2}`

`\mathtt{S_2= (1)/(2) + (1)/(4)}`

`\mathtt{S_2= (2+1)/(4) }`

`\mathbf{S_2= (3)/(4) }`

Therefore, the second partial sum
`\mathbf{S_2= (3)/(4) }`

The third partial sum is :
`\mathtt{S_3= a_1+a_2+a_3}`

`\mathtt{S_3= ((1)/(2))^1 + ((1)/(2))^2+((1)/(2))^3 }`

`\mathtt{S_3= (1)/(2) + (1)/(4)+(1)/(8)}`

`\mathtt{S_3= (4+2+1)/(8)}`

`\mathbf{S_3= (7)/(8)}`

Therefore, the third partial sum
`\mathbf{S_3= (7)/(8)}`

The fourth partial sum :
`\mathtt{S_4= a_1+a_2+a_3+a_4}`

`\mathtt{S_4= ((1)/(2))^1 + ((1)/(2))^2+((1)/(2))^3+((1)/(2))^4 }`

`\mathtt{S_4= (1)/(2) + (1)/(4)+(1)/(8)+(1)/(16)}`

`\mathtt{S_4= (8+4+2+1)/(16)}`

`\mathbf{S_4= (15)/(16)}`

Therefore, the fourth partial sum
`\mathbf{S_4= (15)/(16)}`

The fifth partial sum :
`\mathtt{S_5= a_1+a_2+a_3+a_4+a_5}`

`\mathtt{S_5= ((1)/(2))^1 + ((1)/(2))^2+((1)/(2))^3+((1)/(2))^4 +((1)/(2))^5 }`

`\mathtt{S_5= (1)/(2) + (1)/(4)+(1)/(8)+(1)/(16)+(1)/(32)}`

`\mathtt{S_5= (16+8+4+2+1)/(32)}`

`\mathbf{S_5= (31)/(32)}`

Therefore, the fifth partial sum
`\mathbf{S_5= (31)/(32)}`

The sixth partial sum:
`\mathtt{S_5= a_1+a_2+a_3+a_4+a_5+a_6}`

`\mathtt{S_6= ((1)/(2))^1 + ((1)/(2))^2+((1)/(2))^3+((1)/(2))^4 +((1)/(2))^5 +((1)/(2))^6 }`

`\mathtt{S_6= (1)/(2) + (1)/(4)+(1)/(8)+(1)/(16)+(1)/(32)+(1)/(64) }`

`\mathtt{S_6= (32+16+8+4+2+1)/(64)}`

`\mathbf{S_6= (63)/(64)}`

Therefore, the sixth partial sum
`\mathbf{S_6= (63)/(64)}`