the sum of series
(1×2)+(2×2²)+(3×2³)+(4×2⁴)+...(n×2^n)=?
If u don't know please don't answer
Answers
Answered by
7
well u can add them like this...
(2+2²+2³+2⁴+2⁵+2⁶...+2^n=2(2^(n+1)-1)
+2²+2³+2⁴+2⁵+2⁶...+2^n=4(2^n-1)
+2³+2⁴+2⁵+2⁶...+2^n=8(2^(n-1)-1)
: :
: :
. 2^n=2^n
-------------------------------------------
=n2^{n+1}-(2^(n+1)-2)=(2^(n+1))×(n-1)+2
=
(2+2²+2³+2⁴+2⁵+2⁶...+2^n=2(2^(n+1)-1)
+2²+2³+2⁴+2⁵+2⁶...+2^n=4(2^n-1)
+2³+2⁴+2⁵+2⁶...+2^n=8(2^(n-1)-1)
: :
: :
. 2^n=2^n
-------------------------------------------
=n2^{n+1}-(2^(n+1)-2)=(2^(n+1))×(n-1)+2
=
manitkapoor2:
did i make any mistake?
Answered by
5
Answer:
Step-by-step explanation:
2+2²+2³+2⁴+2⁵+2⁶...+2^n=2(2^(n+1)-1)
+2²+2³+2⁴+2⁵+2⁶...+2^n=4(2^n-1)
+2³+2⁴+2⁵+2⁶...+2^n=8(2^(n-1)-1)
: :
: :
. 2^n=2^n
-------------------------------------------
=n2^{n+1}-(2^(n+1)-2)=(2^(n+1))×(n-1)+2
=2^n-1(n + 1) + 2
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