Happy New Year - 2
Find the possible value of k for which the equation m = 1 ∑ k ( x + m − 1 ) ( x + m ) = 1 0 k has a solution α and α + 1 for some α .
The answer is 11.
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2 solutions
Yeah. There are many problems which are similar to each other throughout brilliant
( y − 1 ) y = 3 3 ( y − 1 ) y = 3 y 3 − ( y − 1 ) 3 − 3 1 ∴ 1 0 k = 3 1 [ m = 1 ∑ k { ( x + m ) 3 − ( x + m − 1 ) 3 } − m = 1 ∑ k 1 ] ⇒ 1 0 k = 3 1 [ ( x + k ) 3 − x 3 − k ] = 3 1 [ 3 k x 2 + 3 k 2 x + k 3 − k ] ⇒ 3 x 2 + 3 k x + k 2 − 3 1 = 0 & ( α − β ) 2 = ( α + β ) 2 − 4 α β = k 2 − 4 3 k 2 − 3 1 = 3 − k 2 + 1 2 4 ⇒ 1 = 3 − k 2 + 1 2 4 ⇒ k = 1 1
Correction k = 1 1
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Thanks,edited.
But why -11 is not the answer?Am i missing something?
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k is the upper bound in the summation. Hence it cannot be negative.
This Problem is same as;
Keep it up ; By....Akshat Sharda,