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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`+...7th term is 167th term is 18Sum of first 10 terms is `(505)/4`Sum of first 10 terms is `(45)/4`A. 7th term is 16B. 7th term is 18C. sum of first 10 terms is `(505)/(4)`D. sum of first 10 terms is `(405)/(4)` |
Answer» Correct Answer - A::C `:.S=1+(1)/(1+3)(1+2)^(2)+(1)/(1+3+5)(1+2+3)^(2)+"......."` `T_(n)=(1)/(1+3+5+7+"........"" n terms ")*(1+2+3+"......."" n terms " )^(2)` `=(1)/([(n)/(2)[2*1+(n-1)*2]])*((n(n+1))/(2))^(2)=((n+1)^(2))/(4)` (a) `T_(7)=((7+1)^(2))/(4)=(64)/(4)=16` (b) `S_(10)sum_(n=1)^(10)((n+1)/(2))^(2)=(1)/(4)sum_(n=1)^(10)(n^(2)+2n+1)` `=(1)/(4)(sum_(n=1)^(10)n^(2)+2sum_(n=1)^(10)n+sum_(n=1)^(10)1)` `=(1)/(4)((10xx11xx21)/(6)+(2xx10xx11)/(2)+10)` `=(1)/(4)(385+110+10)=(505)/(4)` |
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