Question

1^3+2^3+3^3+......+k^3=16900 find 1 +2+3+.....+k?​

Answer 1

The value of 1 +2 +3 +.....+k = 130

Given:

1³ +2³ +3³ +......+k³ = 16900  

To find:

Find the value of 1 +2 +3 +.....+k?​  

Solution:

Formula used:

Sum of cubes of n natural numbers = [n(n + 1)/2]²  

Sum of n natural numbers = [n (n+1)]/2

Given that,

1³ +2³ +3³ +......+k³ = 16900  

By the given formulas,

Sum of cubes of k numbers = [k(k + 1)/2]² ---- (1)

Sum of k numbers = [k (k+1)]/2 ----- (2)

Given that

1³ +2³ +3³ +......+k³ = 16900

=> [k(k + 1)/2]² = 16900

Apply cube root on both sides

=> √[k(k + 1)/2]² = √16900  

=> k(k + 1)/2 = 130 -----(3)

From (2) and (3)

Sum of k numbers = 130

=> 1 + 2 + 3 + .. + k = 130  

Therefore,

The value of 1 +2 +3 +.....+k = 130

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