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1+ (1+3) + (1+3+ 5) + ….n brackets =...

`1+ (1+3) + (1+3+ 5) + ….n` brackets =

A

`(n (n+1) (n+2) )/( 6) `

B

`(n (n +1) (3n^(2) + 23n + 46))/( 12)`

C

`(n(27n^(3) + 90n^(2) + 45n- 50))/( 4)`

D

`(n (n+1) (2n +1) )/( 6)`

Text Solution

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The correct Answer is:
D
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  1. A = sum of first 10 natural numbers, B = sum of squares of first 10 ...

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  2. If sum (r=1)^(n) t(r) = (1)/( 12) n (n+1) (n+2) then sum(r=1)^(n) (1)...

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  3. 1+ (1+3) + (1+3+ 5) + ….n brackets =

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  4. 1+ ((1)/(3) + (1)/(3^2) ) + (( 1)/( 3^3) + (1)/( 3^4) + (1)/( 3^5) ) ...

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  5. Sum of n^( th) bracket of (1) + (2+ 3+4) + ( 5+ 6+ 7+ 8+9) + ….. is

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  6. { n (n+1) (2n+1) : n in Z} sub

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  7. If a,b and n are natural numbers then a^(2n-1) + b^(2n-1) is divisible...

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  8. If 'a' and 'b'are natural numbers such that a^(2) - b^(2) is prime num...

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  9. AA n in N, 49^(n) + 16n -1 is divisible by

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  10. AA n in , 7^(2n) + 3^(n-1) . 2^(3n-3) is divisible by

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  11. Sum of the cubes of three successive natural number is divisible by

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  12. For every natural number n, 3^(2n+2) - 8n -9 is divisible by

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