Below given are two formulae for long summation n1 1 2 3 4 n n n 1 2 n2 1

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Below given are two formulae for long summation.
1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2
2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1)
The value of
(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be: Asked In IES EC A M ANKIT KALLURAYA (10 years ago); Unsolved Read Solution (5); Is this Puzzle helpful? (0) (0) Submit Your Solution Algebra

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There r few things whch we knw whch r nt capable of mathmatical reasoning nd when these can nt it is a sign dt our knowledge of them is very smal nd confusd nd where a mathmaticl reasing can be had . It is as dark a foly to make use a candle stick as to g

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Latest IES EC Aptitude Question SOLUTION: Below given are two formulae for long summation. 1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2 2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1) The value of (1% o

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(#M40106475) IES EC Numerical Ability question series Keep an EYE

Below given are two formulae for long summation.
1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2
2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1)
The value of
(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be

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Latest IES EC Aptitude Question SOLUTION: Below given are two formulae for long summation. 1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2 2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1) The value of (1% o

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(#M40106475) IES EC Numerical Ability question series Keep an EYE

Below given are two formulae for long summation. 1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2 2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1) The value of (1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be

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Below given are two formulae for long summation.
1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2
2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1)
The value of
(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be