find the sum 1/9+6 1/3 please give full answer
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Answer:
T
n
=
3n×3(n+1)
1
=
9(n)(n+1)
1
=
9
1
[
n(n+1)
n+1−n
]
So, 9T
n
=
n
1
−
n+1
1
and S
n
=∑T
n
=T
1
+T
2
+⋯+T
n
So, 9T
1
=1−
2
1
9T
2
=
2
1
−
3
1
⋮
⋮
9T
n
=
n
1
−
n+1
1
on adding , we get
9S
n
=1−
n+1
1
=
n+1
n
So, S
n
=
9(n+1)
n
⇒ Question:
⇒ find the sum 1/9+6 1/3 please give full answer
=============================================================
→ Solution:
Given,
[tex]\frac{1}{9} + 6 \frac{1}{3}[/tex]
To solve the mixed fraction, multiply the denominator i.e. 3 with 6, and add the resulting number with the numerator i.e. 1.
[tex]\frac{1}{9} + \frac{19}{3}[/tex]
Take L.C.M for the denominators i.e. 9 & 3 as they are not equal and the fraction cannot be solved.
L.C.M of 9 & 3 is 9.
In the 9 table, 9 goes in 1 time. so, multiply 1 with both the numerator and denominator, and In the 3 table, 9 goes in 3 times. so, multiply both the numerator and denominator with 3.
[tex]\frac{1 X 1}{9 X 1 } + \frac{19 X 3 }{3 X 3}[/tex]
[tex]\frac{1}{9} + \frac{57}{9}[/tex]
As denominators are same, we can solve the numerators now!
[tex]\frac{1 + 57}{9} = \frac{58}{9}[/tex]
∴ [tex]\frac{1}{9} + 6 \frac{1}{3} = \frac{58}{9}[/tex]
Answer in mixed fraction (if required)
[tex]\frac{58}{9} = 6 \frac{4}{9}[/tex].