2a 2+ 3a2 + 4a constant
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1
Answer:
Let S=1+2a+3a
2
+4a
3
+ a
n−1
Multiply both sides by a, we get
Sa=0+a+2a
2
+3a
3
....n−1)a
n−1
+na
n
Subtract both equations,
S(1−a)=1+a+a
2
+a
3
...a
n−1
−na
n
Clearly above series is G.P
Common ratio =a
S(1−a)=
a−1
1(a
n
−1)
−na
n
S=
(a−1)
2
1−(a
n
)
−
1−a
na
n
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