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junior college 2 | H2 Maths
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Please help with second part
Date Posted: 4 years ago
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“From N + 1 to 2N” is the same as
“From 1 to 2N” minus “From 1 to N”.
You need to obtain the first part and another expression with 2N instead of N. Then subtract the answers.
“From 1 to 2N” minus “From 1 to N”.
You need to obtain the first part and another expression with 2N instead of N. Then subtract the answers.
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
= 1/(1-1+1) - 1/(1+1+1)
+ 1/(2²-2+1) - 1/(2²+2+1)
+ 1/(3²-3+1) - 1/(3²+3+1)
+ ... ... ... ... ... ...
+ 1/[(N-2)²-(N-2)+1] - 1/[(N-2)²+(N-2)+1]
+ 1/[(N-1)²-(N-1)+1] - 1/[(N-1)²+(N-1)+1]
+ 1/(N²-N+1) - 1/(N²+N+1)
= 1/1 - 1/3
+ 1/3 - 1/7
+ 1/7 - 1/13
+ ... ... ... ... ... ...
+ 1/[(N-2)²-(N-2)+1] - 1/[(N-1)² - 2(N-1) +1+(N-2)+1]
+ 1/[(N-1)²-N+2] - 1/(N²-2N+1+N)
+ 1/(N²-N+1) - 1/(N²+N+1)
= 1/1 - 1/3
+ 1/3 - 1/7
+ 1/7 - 1/13
+ ... ... ... ... ... ...
+ 1/[(N-2)²-(N-2)+1] - 1/[(N-1)² - 2N+N+2]
+ 1/[(N-1)²-N+2] - 1/(N²-N+1)
+ 1/(N²-N+1) - 1/(N²+N+1)
= 1/1 - 1/3
+ 1/3 - 1/7
+ 1/7 - 1/13
+ ... ... ... ... ... ...
+ 1/[(N-2)²-(N-2)+1] - 1/[(N-1)² - N+2]
+ 1/[(N-1)²-N+2] - 1/(N²-N+1)
+ 1/(N²-N+1) - 1/(N²+N+1)
= 1 - 1/(N²+N+1)
for n=1 to 2N
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
= 1 - 1/(2N)²+2N+1)
for n=N+1 to 2N
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
EQUALs to
for n=1 to 2N
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
SUBSTRACT
for n=1 to N
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
= [1 - 1/(2N)²+2N+1)]
MINUS
[ 1 - 1/(N²+N+1) ]
= [1 - 1/(2N)²+2N+1)]
- [ 1 - 1/(N²+N+1) ]
= - 1/(4N²+2N+1) + 1/(N²+N+1)
= 1/(N²+N+1) - 1/(4N²+2N+1)
...
= 1/(1-1+1) - 1/(1+1+1)
+ 1/(2²-2+1) - 1/(2²+2+1)
+ 1/(3²-3+1) - 1/(3²+3+1)
+ ... ... ... ... ... ...
+ 1/[(N-2)²-(N-2)+1] - 1/[(N-2)²+(N-2)+1]
+ 1/[(N-1)²-(N-1)+1] - 1/[(N-1)²+(N-1)+1]
+ 1/(N²-N+1) - 1/(N²+N+1)
= 1/1 - 1/3
+ 1/3 - 1/7
+ 1/7 - 1/13
+ ... ... ... ... ... ...
+ 1/[(N-2)²-(N-2)+1] - 1/[(N-1)² - 2(N-1) +1+(N-2)+1]
+ 1/[(N-1)²-N+2] - 1/(N²-2N+1+N)
+ 1/(N²-N+1) - 1/(N²+N+1)
= 1/1 - 1/3
+ 1/3 - 1/7
+ 1/7 - 1/13
+ ... ... ... ... ... ...
+ 1/[(N-2)²-(N-2)+1] - 1/[(N-1)² - 2N+N+2]
+ 1/[(N-1)²-N+2] - 1/(N²-N+1)
+ 1/(N²-N+1) - 1/(N²+N+1)
= 1/1 - 1/3
+ 1/3 - 1/7
+ 1/7 - 1/13
+ ... ... ... ... ... ...
+ 1/[(N-2)²-(N-2)+1] - 1/[(N-1)² - N+2]
+ 1/[(N-1)²-N+2] - 1/(N²-N+1)
+ 1/(N²-N+1) - 1/(N²+N+1)
= 1 - 1/(N²+N+1)
for n=1 to 2N
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
= 1 - 1/(2N)²+2N+1)
for n=N+1 to 2N
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
EQUALs to
for n=1 to 2N
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
SUBSTRACT
for n=1 to N
∑ [ 1/(n²-n+1) - 1/(n²+n+1)]
= [1 - 1/(2N)²+2N+1)]
MINUS
[ 1 - 1/(N²+N+1) ]
= [1 - 1/(2N)²+2N+1)]
- [ 1 - 1/(N²+N+1) ]
= - 1/(4N²+2N+1) + 1/(N²+N+1)
= 1/(N²+N+1) - 1/(4N²+2N+1)
...
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Date Posted:
4 years ago
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