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Question 29 : If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1+2+3+ ... +n ? [TITA]

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Given series - (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280

Isolate 2n terms on one side

(2n + 2n +.... + 2n) + (1 +3 + 5 +.... + 47) = 5280

Odd numbers from 1 to 47 are added in the above series.

Number of terms from 1 to 47 = 24 terms

Therefore, the number of 2n terms = 24

For computing the value of (1 +3 + 5 +.... + 47),

We know 1 = 1^{2}, 1 +3 = 2^{2}, 1 +3 +5 = 3^{2} and so on

So, (1 +3 + 5 +.... + 47) = 24^{2}

So, 2n x 24 + 24^{2 }= 5280

2n x 24 = 5280 - 24^{2}

2n = 220 - 24

n = 110 - 12

n = 98

So, value of 1+2+3+...+98 = = =

Value of 1+2+3+...+98 = 4851

The question is **"If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1+2+3+ ... +n ? [TITA] " **

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