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Find the sum to n terms of the series , 5+11+19+29+41…

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Let `S_(n)=5+11+19+29+.........T_(n)`
Also `S_(n)=5+11+19+......+T_(n-1)+T_(n)`
(1)-(2) gives
`0=5+[6+8+10+…..` up to (n-1) terms)] `-T_(n)`
`T_(n)=5+` [sum to (n-1) terms of A.P] Here a =6. d=2.
`thereforeT_(n)=5+(n-1)/(2).[2a+(n-2)d]=5+(n-1)/(2)[12+(n-2)2]`
`thereforeT_(n)=5+(n-1)/(2)[12+2n-4]=5+(n-1)/(2)[2n+8]=5+(n-1)(n+4)`
`thereforeT_(n)=5+n^(2)+4n-n-4rArrT_(n)=n^(2)+3n+1`
Now `S_(n)=sumT_(n)=sum(n^(2)+3n+1)=sumn^(2)+3sum_(n)+sum1`
`thereforeS_(n)=(n(n+1)(2n+1))/(6)+3(n(n+1))/(2)+n=(n(n+2)(n+4))/(3)`
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