Let a_{1}=b_{1}=1,a_{n}=a_{n-1}+2,and b_{n}=a | vedclass

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(C) Given $a_{1}=1$ and $a_{n}=a_{n-1}+2$,this is an arithmetic progression with first term $1$ and common difference $2$. Thus,$a_{n} = 1 + (n-1)2 =

Vedclass · Accepted Answer

$27560$

Vedclass · Answer

(C) Given $a_{1}=1$ and $a_{n}=a_{n-1}+2$,this is an arithmetic progression with first term $1$ and common difference $2$. Thus,$a_{n} = 1 + (n-1)2 = 2n-1$.Given $b_{1}=1$ and $b_{n}=a_{n}+b_{n-1}$,we have $b_{n} = (2n-1) + b_{n-1}$.Calculating the first few terms: $b_{1}=1$,$b_{2}=3+1=4$,$b_{3}=5+4=9$,$b_{4}=7+9=16$. It follows that $b_{n} = n^{2}$.We need to calculate $\sum_{n=1}^{15} a_{n} b_{n} = \sum_{n=1}^{15} (2n-1)n^{2} = \sum_{n=1}^{15} (2n^{3}-n^{2})$.Using standard summation formulas:$\sum_{n=1}^{N} n^{3} = \frac{N^{2}(N+1)^{2}}{4}$ and $\sum_{n=1}^{N} n^{2} = \frac{N(N+1)(2N+1)}{6}$.For $N=15$:$2 	imes \frac{15^{2} 	imes 16^{2}}{4} - \frac{15 	imes 16 	imes 31}{6} = 2 	imes \frac{225 	imes 256}{4} - \frac{7440}{6} = 28800 - 1240 = 27560$.

Let $a_{1}=b_{1}=1$,$a_{n}=a_{n-1}+2$,and $b_{n}=a_{n}+b_{n-1}$ for every natural number $n \geq 2$. Then $\sum_{n=1}^{15} a_{n} \cdot b_{n}$ is equal to $.........$

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