Find the sum of the series:
Question:

Find the sum of the series:

$(3 \times 8)+(6 \times 11)+(9 \times 14)+\ldots$ to $n$ terms

 

Solution:

In the given question we need to find the sum of the series.

For that, first, we need to find the nth term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is (3 × 8) + (6 × 11) + (9 × 14) + … to n terms.

The series can be written as, [(3 x 1) x (3 x 1 + 5)), (3 x 2) x (3 x 2 + 5))… (3n x (3n + 5))].

So, $\mathrm{n}^{\text {th }}$ term of the series,

$a_{n}=3 n(3 n+5)$

$a_{n}=9 n^{2}+15 n$

Now, we need to find the sum of this series, Sn.

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{n}}$

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(9 \mathrm{n}^{2}+15 \mathrm{n}\right)$

Note:

I. Sum of first $n$ natural numbers, $1+2+3+\ldots n$,

$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$

II. Sum of squares of first $n$ natural numbers, $1^{2}+2^{2}+3^{2}+\ldots n^{2}$,

$\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{2}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$

III. Sum of cubes of first $n$ natural numbers, $1^{3}+2^{3}+3^{3}+\ldots . n^{3}$,

$\sum_{k=1}^{n} k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$

IV. Sum of a constant k, N times,

$\sum_{k=1}^{N} k=N k$

So, for the given series, we need to find,

$\mathrm{S}_{\mathrm{n}}=9 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)+\sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})$

From, the above identities,

$S_{n}=\sum_{n=1}^{n}\left(9 n^{2}\right)+\sum_{n=1}^{n}(15 n)$

$\mathrm{S}_{\mathrm{n}}=9\left(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\right)+15\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)$

$=\left(\frac{n(n+1)}{2}\right)(6 n+18)$

$\mathrm{S}_{\mathrm{n}}=3 \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+3)$

So, Sum of the series, $\mathrm{S}_{\mathrm{n}}=3 \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+3)$

 

 

 

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