Telescope Summation
n = 1 ∑ 2 0 4 n 4 + 1 8 n = 5 8 + 6 5 1 6 + 3 2 5 2 4 + ⋯ + 6 4 0 0 0 1 1 6 0
Let x denote the value of the summation above. Find 8 4 1 x .
The answer is 1680.
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1 solution
Nice application of the Sophie-Germain Identity. (Which caused much pain)
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Ya.... It took me a bit of time to write the series in summation form which was previously given in raw form... :-)
Nice solution i have also applied the same method....
Can you please tell me how can we write any series in summation form ?? Is there any trick for that or just guessing ??
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Its just observation skills and you'll improve it by problem solving only... ;-P
Actually, there are ways by which you can write general term of a list of numbers which follow some kind of pattern by method of differences (There are several other methods as well).
Check out Wikipedia.
Here, we're forunate enough to have the general term already specified. If the question gave only the terms of the series, then it would be a tough task to figure out the general term as it is of degree 4 , which means the 4 th order difference of the terms of series is constant.
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We caluclate the formula for the nth term by differences and by forming AP and GP and then putting it in summation form ? is this you are telling ?
H I N T :The title itself The expression can be converted to general form as: n = 1 ∑ 2 0 4 n 4 + 1 8 n = 8 n = 1 ∑ 2 0 ( 2 n 2 + 1 − 2 n ) ( 2 n 2 + 1 + 2 n ) n = 2 n = 1 ∑ 2 0 2 n 2 + 1 − 2 n 1 − 2 n 2 + 1 + 2 n 1 ( A Telescopic series) = 2 ( 1 − 8 4 1 1 ) = 8 4 1 1 6 8 0 ∴ 8 4 1 1 6 8 0 × 8 4 1 = 1680 Edit: The question is greatly simplified by specifying the series in summation form. Previously it was given in the series form which made the guessing of series in summation form too difficult.