Q. 44.0( 3 Votes )

# If 1^{3< }

Answer :

Given that the series S = 1^{3} + 2^{3} + 3^{3} + … + k^{3} = 8281

Formula to find the sum of first k cubes of natural numbers is

S =

8281 =

Taking square root on both sides,

91 =

k(k + 1) = 182

k^{2} + k = 182

Or k^{2} + k-182 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -182

We get,

k = is invalid because it yields a negative k which doesn’t make sense because number of terms in a series cannot be negative.

= 13

The sum S = 1^{3} + 2^{3} + 3^{3} + ... + k^{3} corresponds to k = 13.

Given series 1 + 2 + 3…. + k, we have k = 13

We have S =

=

=

S = 91

The sum S = 1^{3} + 2^{3} + 3^{3} + ... + k^{3} corresponds to k = 13 and 1 + 2 + 3…. + k = 91.

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