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| 1. |
. find the sum of the first n terms of the series 3 + 7 + 13 + 21 + 31 + ...... |
| Answer» Given: Sn = 3 + 7 + 13 + 21 + 31 + ...... + an-1 + an……….(i)Also Sn = 3 + 7 + 13 + 21 + 31 + ...... + an-2\xa0+ an-1\xa0+ an ……….(ii)Subtracting eq. (i) from eq. (ii), 0 = 3 + ( 4 + 6 + 8 + 10 + ....... up to (n - 1) terms) - an\xa0{tex}\\Rightarrow a _ { n } = 3 + \\frac { n - 1 } { 2 } [ 2 \\times 4 + ( n - 2 ) \\times 2 ]{/tex}{tex}\\Rightarrow a _ { n } = 3 + \\frac { n - 1 } { 2 } [ 8 + 2 n - 4 ]{/tex}{tex}\\Rightarrow{/tex}\xa0an = 3 + (n - 1) (n + 2){tex}\\Rightarrow{/tex}\xa0an = 3 + n2 + n - 2{tex}\\Rightarrow{/tex}\xa0an = n2 + n + 1{tex}\\therefore{/tex}\xa0{tex}{S_n} = \\sum\\limits_{k = 1}^n {{a_{_k}}} = \\sum\\limits_{k = 1}^n {({k^{^2}}} + k + 1){/tex}= (12 + 1 + 1) + (22 + 2 + 1) + (32 + 3 + 1) + ...... +(n2 + n + 1)\xa0= (12 + 22 + 32 + ....... + n2) + (1 + 2 + 3 + ...... + n) + n\xa0{tex}= \\frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \\frac { n ( n + 1 ) } { 2 } + n{/tex}{tex}= n \\left[ \\frac { 2 n ^ { 2 } + 3 n + 1 + 3 n + 3 + 6 } { 6 } \\right]{/tex}{tex}= n \\left[ \\frac { 2 n ^ { 2 } + 6 n + 10 } { 6 } \\right]{/tex}{tex}= \\frac { n } { 3 } \\left( n ^ { 2 } + 3 n + 5 \\right){/tex} | |