Answer

Verified

421.8k+ views

Hint: In any series numbers follow a particular order. Before finding the sum of series, observe the given series to find that order to make simple calculations.

Complete step-by-step answer:

In the given series, the first term is 1.4.7, the second term is 4.7.10 and the third term is 7.10.13 and so on.

We have to find the general form of the series. Observing the given series,

The given series follows an order, such that ‘n’ th term is $${T_n} = (3n - 2) \cdot (3n + 1) \cdot (3n + 4)$$

$$ \Rightarrow {T_n} = 27{n^3} + 27{n^2} - 18n - 18$$

Now we got a generalized form for the given series, we need to find the sum of the series.

Taking summation over ‘n’ on both sides of $${T_n}$$

$$ \Rightarrow \sum {{T_n} = \sum {(27{n^3} + 27{n^2} - 18n - 8} } )$$

$$ \Rightarrow \sum {{T_n} = \sum {(27{n^3}) + \sum {(27{n^2})} - \sum {(18n)} - \sum {(8)} } } $$

$$ \Rightarrow \sum {{T_n} = 27\sum {({n^3}) + 27\sum {({n^2})} - 18\sum {(n)} - 8} } n$$

$$\left[ {\because \sum {{n^3}} = \dfrac{{{n^2}{{(n + 1)}^2}}}{4};\sum {{n^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6};\sum {n = \dfrac{{n(n + 1)}}{2}} ;\sum {k = n \times k} } \right]$$

$$ \Rightarrow \sum {{T_n} = 27\left( {\dfrac{{{n^2}{{(n + 1)}^2}}}{4}} \right) + 27\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right) - 18\left( {\dfrac{{n(n + 1)}}{2}} \right)} - 8n$$

$$ \Rightarrow \sum {{T_n} = n(n + 1)\left[ {\dfrac{{27n(n + 1)}}{4} + \dfrac{{27n(2n + 1)}}{7} - \dfrac{{18}}{2}} \right]} - 8n$$

$$ \Rightarrow \sum {{T_n} = n(n + 1)\left[ {\dfrac{{27({n^2} + n)}}{4} + \dfrac{{27(2{n^2} + n)}}{7} - 9} \right]} - 8n$$

$\therefore $The sum of the given series is $$\sum {{T_n} = n(n + 1)\left[ {\dfrac{{27({n^2} + n)}}{4} + \dfrac{{27(2{n^2} + n)}}{7} - 9} \right]} - 8n$$

Note: First we found the general form for the given series. We used basic summation formulae to get the required solution. We applied those formulae and simplified the equation.

$$\sum {k = n \times k} $$, Where k is a constant.

Complete step-by-step answer:

In the given series, the first term is 1.4.7, the second term is 4.7.10 and the third term is 7.10.13 and so on.

We have to find the general form of the series. Observing the given series,

(3n-2) | (3n+1) | (3n+4) | |

n=1 | 1 | 4 | 7 |

n=2 | 4 | 7 | 10 |

n=3 | 7 | 10 | 13 |

The given series follows an order, such that ‘n’ th term is $${T_n} = (3n - 2) \cdot (3n + 1) \cdot (3n + 4)$$

$$ \Rightarrow {T_n} = 27{n^3} + 27{n^2} - 18n - 18$$

Now we got a generalized form for the given series, we need to find the sum of the series.

Taking summation over ‘n’ on both sides of $${T_n}$$

$$ \Rightarrow \sum {{T_n} = \sum {(27{n^3} + 27{n^2} - 18n - 8} } )$$

$$ \Rightarrow \sum {{T_n} = \sum {(27{n^3}) + \sum {(27{n^2})} - \sum {(18n)} - \sum {(8)} } } $$

$$ \Rightarrow \sum {{T_n} = 27\sum {({n^3}) + 27\sum {({n^2})} - 18\sum {(n)} - 8} } n$$

$$\left[ {\because \sum {{n^3}} = \dfrac{{{n^2}{{(n + 1)}^2}}}{4};\sum {{n^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6};\sum {n = \dfrac{{n(n + 1)}}{2}} ;\sum {k = n \times k} } \right]$$

$$ \Rightarrow \sum {{T_n} = 27\left( {\dfrac{{{n^2}{{(n + 1)}^2}}}{4}} \right) + 27\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right) - 18\left( {\dfrac{{n(n + 1)}}{2}} \right)} - 8n$$

$$ \Rightarrow \sum {{T_n} = n(n + 1)\left[ {\dfrac{{27n(n + 1)}}{4} + \dfrac{{27n(2n + 1)}}{7} - \dfrac{{18}}{2}} \right]} - 8n$$

$$ \Rightarrow \sum {{T_n} = n(n + 1)\left[ {\dfrac{{27({n^2} + n)}}{4} + \dfrac{{27(2{n^2} + n)}}{7} - 9} \right]} - 8n$$

$\therefore $The sum of the given series is $$\sum {{T_n} = n(n + 1)\left[ {\dfrac{{27({n^2} + n)}}{4} + \dfrac{{27(2{n^2} + n)}}{7} - 9} \right]} - 8n$$

Note: First we found the general form for the given series. We used basic summation formulae to get the required solution. We applied those formulae and simplified the equation.

$$\sum {k = n \times k} $$, Where k is a constant.

Recently Updated Pages

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many crores make 10 million class 7 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Explain zero factorial class 11 maths CBSE