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Question

If tn=14(n+2)(n+3) for n=1,2,3.... then 1t1+1t2+1t3+.........+1t2003=

A
40063006
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B
40033007
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C
40063008
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D
40063009
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Solution

The correct option is D 40063009
tn=(n+2)(n+3)4
1tn=4(n+2)(n+3)
=4[1n+21n+3]
Now sum =1t1+1t2+1t3+...+1t2002+1t2003
=4[1314+1415+15+16+.....+1200412005+1200512006]
=4[1312006]
=4[200633(2006)]
=2(2003)3009=40063009

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