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  • 3 answers

Thufail Ahamed 5 hours ago

Study

Divya Yaduvanshi 13 hours ago

Believe on yourself and focus on knowledge

Aadika 😍 13 hours ago

Lots of practice only. Passing marks : 27 in cbse ....
  • 5 answers

Ram Lal Singh 7 hours ago

4536 correct answer

Mukul Mittal 18 hours ago

4536 correct

Sikandar Kumar 1 day, 11 hours ago

9×9×8×7=4536

Shubham Kumar 1 day, 12 hours ago

9×9×8×7=4536 If u are taking. 0,1,2,3,4,5,6,7,8,9

Ashwin Patel 1 day, 14 hours ago

24
  • 1 answers

Shubham Kumar 1 day, 22 hours ago

Is this a real question or u made it 😅👦🏻
  • 6 answers

Aadika 😍 1 day, 12 hours ago

Maths ka blue print mere friend ko chahiye tha ...

Aadika 😍 1 day, 12 hours ago

No, I have chosen Hindi

Shubham Kumar 1 day, 12 hours ago

Aadika u have taken both maths and hindi?

Aadika 😍 2 days, 1 hour ago

Jaldii help karo

Shubham Kumar 2 days, 1 hour ago

U r a kvaian so it might help

Shubham Kumar 2 days, 1 hour ago

I got papers
  • 1 answers

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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  • 1 answers

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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  • 1 answers

Sia 🤖 2 days, 18 hours 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}

  • 1 answers

Subham Panwar 3 days, 20 hours ago

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  • 0 answers
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  • 6 answers

Vaibhav Kumar 12 hours ago

Answer is not 1

Vaibhav Kumar 12 hours ago

All you are wrong

Adarsh Kesharwani 3 days, 15 hours ago

1

Atharv Baranwal 4 days, 19 hours ago

1

Pruthviraj Gurav 5 days, 14 hours ago

1

Aditya Narayan Singh 5 days, 14 hours ago

1

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