# Let Sn denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then ∑r=1nSrsr equals ______. - Mathematics

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Let Sn denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then sum_(r = 1)^n S_r/s_r equals ______.

#### Options

• (n(n + 1)(n + 2))/6

• (n(n + 1))/2

• (n^2 + 3n + 2)/2

• None of these

#### Solution

Let Sn denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then sum_(r = 1)^n S_r/s_r equals (n(n + 1)(n + 2))/6.

Explanation:

Given that sum_(i = 1)^n S_r/s_r = S_1/s_1 + S_2/s_2 + S_3/s_3 + ... + S_n/s_n

Let Tn be the nth term of the above series

∴ Tn = S_n/s_n

= ([(n(n + 1))/2]^2)/((n(n + 1))/2)

= (n(n + 1))/2

= (n^2 + n)/2

Now sum of the given series

sum"T"_"n" = 1/2 sum [n^2 + n]

= 1/2 [sum n^2 + sum n]

= 1/2 [(n(n + 1)(2n + 1))/6 + (n(n + 1))/2]

= 1/2 * (n(n + 1))/2 [(2n + 1)/3 + 1]

= (n(n + 1))/4 [(2n + 1 + 3)/3]

= (n(n + 1))/4 * ((2n + 4))/3

= (n(n + 1)(n + 2))/6

Concept: Sum to N Terms of Special Series
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 9 Sequences and Series
Exercise | Q 24 | Page 163