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Ask QuestionPosted by Dipesh Nagar 6 years, 2 months ago
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Posted by Yuvraj Yadav 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
Given: Sn = 3 + 7 + 13 + 21 + 31 + ...... + an-1 + an……….(i)
Also Sn = 3 + 7 + 13 + 21 + 31 + ...... + an-2 + an-1 + an ……….(ii)
Subtracting eq. (i) from eq. (ii), 0 = 3 + ( 4 + 6 + 8 + 10 + ....... up to (n - 1) terms) - an
{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} an = 3 + (n - 1) (n + 2)
{tex}\Rightarrow{/tex} an = 3 + n2 + n - 2
{tex}\Rightarrow{/tex} an = n2 + n + 1
{tex}\therefore{/tex} {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)
= (12 + 22 + 32 + ....... + n2) + (1 + 2 + 3 + ...... + n) + n
{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}
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Yuvraj Singh 6 years, 1 month ago
0Thank You