The sum 1 + 2 ⋅ 3 + 3 ⋅ 32 + …. + 10 ⋅ 39 is equal to
- \(\frac{2.3^{12}+10}{4}\)
- \(\frac{19.3^{10}+1}{4}\)
- \(5.5^{10}-2\)
- \(\frac{9.3^{10}+1}{2}\)
The Correct Option is B
Solution and Explanation
Let
\(s=1.3^o+2.3^1+3.3^2+.......+10.3^9\)
\({3s=1.3^1+2.3^2+.......+10.3^{10}}\)
\(-2s=(1.3^o+1.3^1+1.3^2+.......+1.3^9)-10.3^{10}\)
\(⇒s=\frac{1}{2}[10.3^{10}-\frac{3^{10}-1}{3-1}]\)
\(⇒s=\frac{1}{2}[\frac{20.3^{10}{{-3^{10}+1}}}{2}]\)
\(⇒S=\frac{19.3^{10}+1}{4}\)
So, The correct option is(B): \(\frac{19.3^{10}+1}{4}\)
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Sum to n Terms of Special Series
A sequence is a list of numbers in a certain or particular order. Each number in a sequence is called a term. A series is the sum of all the terms of a given sequence is called a series. A finite series with a countable number of terms is commonly known as a finite series, and that with an infinite number of terms is called an infinite series. The sum to n terms of a series is reflected by Sn.
In mathematics, we may come across distinct types of series such as geometric series, arithmetic series, harmonic series, etc. Apart from these, we can notice some special series for which we can find the sum of the terms using distinct techniques.
Some special series are given below:
- Sum of first n natural numbers = 1 + 2 + 3 +…+ n
- Sum of squares of the first n natural numbers = 12 + 22 + 32 +…+ n2
- Sum of cubes of the first n natural numbers = 13 + 23 + 33 +…+ n3