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For the series, S=1+1/((1+3))(1+2)^2+1/(...

For the series, `S=1+1/((1+3))(1+2)^2+1/((1+3+5))(1+2+3)^2+1/((1+3+5+7)) (1+2+3+4)^2` +...

A

`7^(th) " term is " 16`

B

`7^(th) " term " is 18`

C

Sum of first 10 terms is `(505)/(4)`

D

Sum of first 10 terms is `(405)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`S=1+1/((1+3))(1+2)^(2)+1/((1+3+5))(1+2+3)^(2)+1/((1+3+5+7))(1+2+3+4)^(2)+…`
The `r^(th)` term is given by
`T_(r)=1/r^(2)(1+2+..+r)^(2)=1/r^(2){(r(r+1))/2}^(2)=(r^(2)+2r+1)/4`
`thereforeT_(7)=16`
and `S_(10)=sum_(r=1)^(10)T_(r)`
=`1/4{((10)(10+1)(20+1))/6+(10)(10+1)+10}`
`=505/4`
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CENGAGE-PROGRESSION AND SERIES-Exercise (Multiple & Comprehension)
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