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Sreekutty N 1 day, 18 hours ago

4th term we should find it by using general formula Tr+1 = ^nCr X^n-r Y^r
r=3 , n=12 ,X=x y=2y. Substute it in the formula .

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Samiksha Bhardwaj 2 days, 18 hours ago

Tan70=tan(20+50)
Nd using the identity of tan(A+B)
Tan70(1-tan20.tan50)=tan20+tan50
Tan70-tan20.tan70.tan50=tan20+tan50
And..tan70.tan20=tan70cot70=1
Now
Tan70-tan50=tan20+tan50
Or
Tan70=tan20+2tan50

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Sia 🤖 2 days, 18 hours ago

Given: S_{n} = 3 + 7 + 13 + 21 + 31 + ...... + a_{n-1} + a_{n}……….(i)

Also S_{n} = 3 + 7 + 13 + 21 + 31 + ...... + a_{n-2} + a_{n-1} + a_{n} ……….(ii)

Subtracting eq. (i) from eq. (ii), 0 = 3 + ( 4 + 6 + 8 + 10 + ....... up to (n - 1) terms) - a_{n}

{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} a_{n} = 3 + (n - 1) (n + 2)

{tex}\Rightarrow{/tex} an = 3 + n^{2} + n - 2

{tex}\Rightarrow{/tex} a_{n} = n^{2} + 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}

= (1^{2} + 1 + 1) + (2^{2} + 2 + 1) + (3^{2} + 3 + 1) + ...... +(n^{2} + n + 1)

= (1^{2} + 2^{2} + 3^{2} + ....... + n^{2}) + (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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