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Find the value of lambda the equation is...

Find the value of `lambda` the equation is `2x+3y+lambda` where `x=2` and `y=1`

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The correct Answer is:
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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