Mathematics Class XI

Question 1 of 25

1. Question
1 point(s)
$n(n+1) (n+5) $ is multiple of _______ ,$ \forall n\in N $
- 4
- 3
- 5
- 7
Correct

Incorrect

Hint

$f(n)= n(n+1)(n+5)$
For $n=1, f(1)= 1(1+1)(1+5) =1 \times 2 \times 6 = 12$ , is a multiple of $3$ .
For $n=2, f(2)= 2(2+1)(2+5) =2 \times 3 \times 7 = 42$ , is a multiple of $3$ .
For $n=k, f(k)= k(k+1)(k+5)$ is true.
For $n=k+1,$ $\begin{aligned} f(k+1)&=(k+1)(k+1+1) (k+1+5)\\ \\ &=(k+1)(k+2) (k+6) \\ \\ &=(k+1)(k+2)k+6(k+1)(k+2) \\ \\ &=k(k+1)(k+2)+6(k+1)(k+2) \\ \\& =k(k+1)(k+5-3)+6(k+1)(k+2) \\ \\ &=k(k+1)(k+5)-3k(k+1)+6(k+1)(k+2) \\ \\ &=k(k+1)(k+5)+3(k+1)+(k+4) \\ \\ &=3m +3(k+1)(k+4) \\ \\ &=3[m+(k+1)(k+4)]\end{aligned}$ , is multiple of $3$
$f(k+1)$ is true.
$f(n)$
Question 2 of 25

2. Question
1 point(s)
Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing $ n $ shots.
- $\dfrac{n(n+1)(n+2)}{3} $
- $\dfrac{n(n+1)(n+2)}{6} $
- $\dfrac{n(n+2)}{6} $
- None of these
Correct

Incorrect

Hint

The number of shots in the lower layer $\begin{aligned} &=n+(n-1)+ (n-2)+...+1\\&=\dfrac{n(n+1)}{2}\\&=\dfrac{n^2+n}{2}\end{aligned}$
.
We obtain the number of shots in 2nd, 3rd,…layer by written $(n-1), (n-2)$
,… respectively.
So, total shots $\begin{aligned} &=\sum\left ( \dfrac{n^2+n}{2} \right )\\&=\dfrac{1}{2}\sum (n^2+n)\\&=\dfrac{1}{2}\left [ \dfrac{n(n+1)(2n+1)}{6} +\dfrac{n(n+1)}{2}\right ]\\&=\dfrac{1}{6}\left [ n(n+1)(n+2) \right ]\end{aligned}$ .
Question 3 of 25

3. Question
1 point(s)
$(n^2+n) $ is _ ,$ \forall n\in N $
- Even
- Odd
- Either even or odd
- None of these
Correct

Incorrect

Hint

$f(n)=n^2+n$
For $n=1, f(1)=1^2+1 =2$ even.
For $n=2, f(2)=2^2+2 =6$ even.
For $n=k, f(k)=k^k+k$ even.
For $n=(k+1),$
$\begin{aligned} f(k+1)&=(k+1)^2+(k+1)\\&=(k+1)^2+(k+1)\\&=k^2+2k+1+k+1\\&=k^2+3k+2\\&=(k^2+k)+(2k+2)\\&=(2m)+2(k+1)\\&=2(m+k+1)\end{aligned}$ , is even.
$f(k+1)$ is true.

$\therefore f(n)=n^2+n$ , is even, $\forall n\in N$ .
Question 4 of 25

4. Question
1 point(s)
For all $n\in N $, $3 \times 5^{2n+1}+ 2^{3n+1} $ is divisible by__________.
- 19
- 17
- 23
- 25
Correct

Incorrect

Hint

$f(n)= 3 \times5 ^{2n+1} +2^{3n+1}$
For $n=1, \begin{aligned} f{(1)}&=3 \times 5^3+2^4=3 \times 125 +16\\&=375+16 =391\end{aligned}$ , is divisible by $17$ .
For $n=2, \begin{aligned} f{(2)}&=3 \times 5^5+2^7=3 \times 3125+128=9503\end{aligned}$ , is divisible by $17$ .
$\therefore$ For $n=k, f(k) = 3\times 5 ^{2k+1}+2 ^{3k+1}$ is true.
For $n=k+1$
$\begin{aligned} f(k+1)&=3 \times 5^{2(k+1)+1}+2^{3(k+1)+1}\\&=3\times 5^{2k+1}\cdot 5^2+2^{3k+1}\cdot 2^3\\&=3\times5^{2k+1} \cdot 5^2+2^{3k+1} \cdot5^2-2^{3k+1}\cdot5^2+2^{3k+1}\cdot2^3\\&=5^2 (3\times 5^{2k+1}+2^{3k+1})+2^{3k+1}(-25+8)\\&=5^2(17m)+2^{3k+1}(-17)=17[25m-2^{3k+1}]\end{aligned}$ is divisible by $17$ .
$\therefore$ $f(k+1$ is true.
$\therefore$ $f(n)$ is true, $\forall n \in N$ .
Question 5 of 25

5. Question
1 point(s)
$f(n)= 12^n +25^{n-1} $ is divisible by ____________.
- 13
- 11
- 12
- 17
Correct

Incorrect

Hint

$f(n)= 12^n +25^{n-1}$
For $n=1,f(1)= 12^1 +25^{1-1}=12+1=13$
For $n=2,f(2)= 12^2 +25^{2-1}=144+25=169=13\times 13$
$\therefore$ For $n=k,f(k)=12^k+25^{k-1}$ is divisible by $13$ .
For $n=k+1,$
$\begin{aligned} f(k+1)&=12^{k+1}+25 ^{k+1-1}\\&=12^k\cdot12+25^{k+1}\cdot 25\\&=12^k(13-1)+25^{k-1}(26-1)\\&=13\cdot 12^k+26\cdot25^{k-1}-12^k-25^{k-1}\\&=13\left ( \dfrac{12^k+2\cdot25^{k-1}}{a}\right )- \left ( \dfrac{12^k+25^{k-1}}{13m} \right )\\ \\&=13\times a-13m =13(a-m)\end{aligned}$ is divisible by $13$ .
$\therefore$ $f(k+1)$ is true.
$\therefore$ $f(n)$ is true, $\forall n\in N$ .
Question 6 of 25

6. Question
1 point(s)
$f(n)= 11^{n+2}+12^{2n+1} $ is divisible by______________.
- 121
- 131
- 133
- 101
Correct

Incorrect

Hint

$f(n)= 11^{n+2}+12^{2n+1}$
For $n=1,\begin{aligned} f(1)&=11^3+12^3\\&=(11+12)(11^2-11\times12+12^2)\\&=23\times133\end{aligned}$ , is divisible by $133$ .
For $n=k,f(k)=11^{k+2}+12^{2k+1}$ , is divisible by $133$ .
For $n=k+1,$
$\begin{aligned}f(k+1)&=11^{k+1+2}+12^{2(k+1)+1}\\&=11^{k+3}+12^{2k+3}\\&=11\cdot11^{k+2}+12^2\cdot12^{2k+1}\\&=11\cdot11^{k+2}+(11+133)12^{2k+1}\\&=11\cdot11^{k+2}+11\cdot12^{2k+1}+133\times 12^{2k+1}\\&=11(11^{k+2}+12^{2k+1})+133\times 12^{2k+1}\\&=11 \times 133m+133\times12^{2k+1}\\&=133(11m+12^{2k+1})\end{aligned}$ , is divisible by $133$ .
$\therefore$ $f(k+1 )$ is true.
$\therefore$ $f(n)$ is true, $\forall n\in N$ .
Question 7 of 25

7. Question
1 point(s)
$f(n)= 2 \times 7^n +3\times5^n-5 $ is divisible by ___________.
- 23
- 24
- 26
- 25
Correct

Incorrect

Hint

$f(n)= 2 \times 7^n +3\times5^n-5$
For $n=1,f(1)= 2\times 7+3\times5-5=24$
For $n=2,f(2)= 2\times 7^2+3\times5^2-5=98 +75-5=168=24\times 7$
For $n=k, f(k)= 2\times 7^k+3\times5^k-5$ is divisible by $24$ .
$\therefore$ For $n=k+1,$
$\begin{aligned} f(k+1)&=2\times7^{k+1}+3\times 5^{k+1}-5\\&=2\times 7^k \times7+3\times5^k\times5-5\\&=2\times 7^k (6+1)+3\times5^k (4+1)-5\\&=12(7^k+5^k)+(2\cdot 7^k+3\cdot5^k-5)\end{aligned}$
Since $7^k$ and $5^k$ are always odd, then $7^k +5^k$ is always even.
$=(12 \times 2m)+24n =24(m+n)$ , is divisible by $24$ .
$\therefore$ $f(k+1)$ is true.
$\therefore$ $f(n)$ is true, $\forall n\in N$ .
Question 8 of 25

8. Question
1 point(s)
For $ \forall n\in N $,$ f(n)=41^n-14^n $ is divisible by___________.
- 23
- 41
- 27
- 14
Correct

Incorrect

Hint

$f(n)=41^n-14^n$
For $n=1, f(1)= 41^1-14^1=27$
For $n=k, f(k)= 41^k-14^k$ is true.
$\therefore$ For $n=k+1,$
$\begin{aligned} f(k+1)&=41^{k+1}-14^{k+1}\\&=41^{k+1}-14^{k+1}\\&=41^k\cdot41-14^{k+1}\\&=41^k(27+14)-14^{k+1}\\&=41^k\times 27+14\times41^k-14^k\times14\\&=27(41^k)+14(41^k-14^k)\end{aligned}$
$14(41^k-14^k)$ is a multiple of $27$ .
$\begin{aligned} &=27(41^k)+14(27m)\\&=27(41^k+14m)\end{aligned}$ , is divisible by $27$ .
$\therefore$ $f(k+1)$ is true.
$\therefore$ $f(n)$ is true, $\forall n\in N$ .
Question 9 of 25

9. Question
1 point(s)
$ 2^{3n}-1 $ is divisible by __________.
- 6
- 5
- 9
- 7
Correct

Incorrect

Hint

$f(n)=2^{3n}-1$ , For $n=1,f(1)=2^3-1=7$
For $f(k)$ where $n=k,f(k)=2^{3k}-1$ is true.
$\therefore$ For $n=k+1,$
$\begin{aligned} f(k+1)&=2^{3(k+1)}-1\\&=2^{3k+3}-1\\&=2^{3k}\cdot2^3-1\\&=2^{3k}(7+1)-1\\&=7\cdot2^{3k}+2^{3k}-1\\&=7\cdot2^{3k}+7m\\&=7(2^{3k}+m)\end{aligned}$ is divisible by $7$ .
$\therefore$ $f(k+1)$ is true.
$\therefore$ $f(n)=2^{3n}-1$ is divisible by $f ,\forall n\in N$ .
Question 10 of 25

10. Question
1 point(s)
$ 4^n+15n-1 $ is divisible by ____________.
- 9
- 5
- 7
- 8
Correct

Incorrect

Hint

$f(n)=4^n+15n-1$
For $n=1,f(1)=4+15-1=18$ ,is divisible by $9$ .
For $n=2,f(2)=4^2+15\times2-1=45$ , is divisible by $9$ .
$\therefore$ For $n=k,f(k)=4^k+15k-1$ is true.
$\therefore$ For $n=k+1,$
$\begin{aligned} f(k+1)&=4^{k+1}+15(k+1)-1\\&=4^{k+1}+15(k+1)-1\\&=4^k\cdot4+15k+15-1 \\&=4^k(9-5)+(90k-75k)+(9+5)\\&=9(4^k+10k+1)-5(4^k+15k-1)\\&=9(4^k+10k+1)-5 (9m)\\&=9(4^k+10k+1-5m)\end{aligned}$ , is divisible by $9$ .
$\therefore$ $f(k+1)$ is true.
$\therefore$ $f(n)=4^n+15n-1$ is divisible by $9$ .
Question 11 of 25

11. Question
1 point(s)
$ x^n-y^n $ is divisible by $ (x+y) $ when _____________.
- $n\in Z $
- $ n\in N $
- $n=2a,a\in N $
- $n=2a+1,a\in N $
Correct

Incorrect

Hint

$f(n)= x^n-y^n$
For $n=2, f(2)= x^2-y^2 =(x-y)(x+y)$ , is divisible by $(x+y)$ .
For $n=k, f(k)= x^k-y^k$ is true.
For $n=k+1,$
$\begin{aligned} f(k+2)&=x^{k+2}-y^{k+2}\\&=x^{k+2}-y^{k+2}\\&=x^{k+2}-x^2\cdot y^k+x^2y^k-y^{k+2}\\&=x^2(x^k-y^k)+y^k(x^2-y^2)\\&=x^2\cdot(x+y)m+y^k(x+y)(x-y)\\&=(x+y)[x^2m+y^k(x-y)]\end{aligned}$
$\therefore$ $f(k+2)$ is true.
$\therefore$ $f(n) =x^n-y^n$ is divisible by $( x+y)$ when $n$ is even.
Question 12 of 25

12. Question
1 point(s)
$f(n)=n^3+ (n+1)^3+(n+2)^3 $ is divisible by ___________.
- 3
- 9
- 7
- 5
Correct

Incorrect

Hint

$f(n)=n^3+ (n+1)^3+(n+2)^3$
For $n=1, f(1)=1+2^3+3^3=36$ , is divisible by $9$ .
For $n=k, f(k)=k^3+(k+1)^3+(k+2)^3$ is true.
$\therefore$ For $n=k+1,$
$\begin{aligned} f(k+1)&=(k+1)^3+(k+1+1)^3+(k+1+2)^3\\&=(k+1)^3+(k+2)^3(k+3)^3\\&=(k+1)^3+(k+2)^3+k^3+9k^2+27k+27\\&=[k^3+(k+1)^3+(k+2)^3]+9[k^2+3k+3]\\&=9m+9(k^2+3k+3)\\&=9(m+k^2+3k+3)\end{aligned}$
$\therefore$ $f(k+1)$ is divisible by $9$ .
$\therefore$ $f(n)=n^3+(n+1)^3+(n+2)^3$ is divisible by $9$ .
Question 13 of 25

13. Question
1 point(s)
For $n\in N, 3^{3n} -26n-1 $,is divisible by __________.
- 627
- 676
- 547
- 349
Correct

Incorrect

Hint

$f(n)=3^{3n}26n-1$
$f(n)=3^{3}-26-1=0$
$f(2)=3^{6}-26\times 2-1 = 729-53=676$
$\therefore$ $f(k)=3^{3k}-26k-1$ is true.
$\therefore$ $f(k+1)=3^{3(k+1)}-26(k+1)-1$ is true by induction hypothesis.
$\therefore$ $f(n)=3^{3n}-26n-1$ is divisible by $676$ .
Question 14 of 25

14. Question
1 point(s)
For all $n\in N, n(n+1)(2n+1) $ is divisible by _________.
- 30
- 13
- 6
- 11
Correct

Incorrect

Hint

$f(n)= n(n+1)(2n+1)$
For $n=1, f(1)= 1\times (1+1)\times(2+1)=1\times2\times3 =6$
For $n=2, f(2)= 2\times3 \times 5= 30$ , is divisible by $6$ .
$\therefore$ For $n=k,f(k)= k(k+1)(2k+1)$ is true.
For $n=k+1,$
$\begin{aligned} f(k+1)&=(k+1)(k+1+1)[2(k+1)+1]\\&=(k+1)(k+2)(2k+3)\\&=(k+1)(k+2)(2k+1+2)\\&=(k+1)[k(2k+1+2)+2(2k+1+2)]\\&=(k+1)k(2k+1)+[2k+2(2k+3)](k+1)\\&=6m+(k+1)(6k+6)\\&=6m+6(k+1)(k+1)\\&=6[m+(k+1^2)]\end{aligned}$ , is divisible by $6$ .
$\therefore$ $f(k+1)$ is true.
$\therefore$ $f(n)= n(n+1) (2n+1)$ is true, $\forall n\in N$ .
Question 15 of 25

15. Question
1 point(s)
$ f(n)=2^{3n}-7^n-1, \forall n\in N $ is divisible by ____________.
- 49
- 36
- 45
- 51
Correct

Incorrect

Hint

$f(n)=2^{3n}-7^n-1$
For $n=1, f(1)=2^3-7-1=0$
For $n=2, f(2)=2^6-14-1=64-15=49$
For $n=k, f(k)= 2^{3k} -7k-1$ is true.
$\therefore$ By induction hypothesis it is true for $n=k+1$ .
$\therefore$ $f(k+1)$ is true.
$\therefore$ $f(n)$ is true, $\forall n\in N$ .
For $n=k, f(k)= 2^{3k}-7k-1$ is true.
Question 16 of 25

16. Question
1 point(s)
The number of terms in the expansion of $(n+y+z)^n $ is ___________.
- $\dfrac{n(n+2)}{2} $
- $\dfrac{(n+1)(n+2)}{2} $
- $ \dfrac{(n+1)(n+3)}{2} $
- $ \dfrac{n(n+2)}{2} $
Correct

Incorrect

Hint

Number of terms $=^{n+n-1}Cn$
For $(x+y+z)^n, n=3$
Hence number of terms are
$^{n+3-1}Cn=^{n+2}Cn= \dfrac{(n+2)!}{n!2!}= \dfrac{(n+1)(n+2)}{2}$
Question 17 of 25

17. Question
1 point(s)
A student asked to prove a statement $ P(n) $ by method of induction. He proved $ P(k+1) $ is true whenever $ P(k) $ is true $ \forall k \geq 5,K \in N $ and $ P(5) $ is true. On this basis he could conclude that $ P(n) $ is true_________.
- $ \forall n > 5 $
- $ \forall n$ < 5
- $\forall n \geq 5 $
- None of these
Correct

Incorrect

Hint

If $P(a)$ is true and truth of $P(n)$ implies that $P(n+1)$ is also true, we can conclude that $P(n)$ is true $\forall n \geq a$ , where $n \in N$ .
Hence, we can conclude that $P(n)$ is true $\forall n \geq 5$ .
Question 18 of 25

18. Question
1 point(s)
$ 1\cdot 2\cdot 3 +2\cdot 3\cdot 4 + 3\cdot 4 \cdot 5+… $ up to $ n $ terms is equal to ______________.
- $ \dfrac{1}{4} (n+1)(n-1)(n+2)(n+3) $
- $ \dfrac{1}{4} (n+1)(n+2)(n+3) $
- $\dfrac{1}{4} n(n+1)(n-1)(n+2)(n+3) $
- None of these
Correct

Incorrect

Hint

nth term $=n (n+1)(n+2)$ , where $n =1,2,3,...$
$=n(n^2+3n+2 )= n^3+3n^2+2n$
$\text{Sum} = \dfrac{n(n+1)}{2}$ , if nth term $=n$ .
$=\dfrac{n(n+1)(2n+1)}{6}$ , if nth term $=n^2$ .
$=\dfrac{n^2(n+1)^2}{4}$ , if nth term $=n^3$ .
Hence the required sum
$\begin{aligned} &=\dfrac{n^2(n+1)^2}{4}+\dfrac{3n(n+1)(2n+1)}{6} +\dfrac{2n(n+1)}{2}\\&=\dfrac{n^2(n+1)^2}{4}+\dfrac{n(n+1)(2n+1)}{2}+n(n+1)\\&=n(n+1)[\dfrac{n(n+1)}{4}+\dfrac{2n+1}{2}+1]\\&=n(n+1)[\dfrac{n^2+n+4n+2+4}{4}]\\&=\dfrac{1}{4}n(n+1)(n^2+5n+6)\\&=\dfrac{1}{4}n(n+1)(n+2)(n+3)\end{aligned}$
Question 19 of 25

19. Question
1 point(s)
Consider the statement $ P(n) : “n^2 \geq 100” $, hence $ P(n)\Rightarrow P(n+1) $, for some $ n $, then_________.
- $P(n) $ is true for all $n \geq 2 $
- $P(n) $ is true for all $n \geq 3 $
- $P(n) $ is true for all $n \geq 5 $
- $P(n) $ is true for all $n \geq 10 $
Correct

Incorrect

Hint

$P(n): "n^2 \geq 100"$
For $n=10, n^2 =100$
$n=11,n^2 =121$
$\therefore$ $n^2 \geq 100, \forall n \geq 10$
Question 20 of 25

20. Question
1 point(s)
$P(n): n^2-n+41 $ is a Prime Number , then _________.
- $ P(3) $ is not true.
- $ P(5) $ is not true.
- $ P(41) $ is not true.
- $ P(1) $ is not true.
Correct

Incorrect

Hint

$P(n): n^2-n+41$
$P(1): 1-1+41 =41$ , is a prime number.
$P(3): 9-3+41 =47$ , is prime number.
$P(5):25-5+41 =61$ , is a prime number.
$P(4):41^2-41+41 =41^2$ , is not a prime number.
$\therefore$ $P(41)$ is not true for $n^2-n+41$ is a prime number.
Question 21 of 25

21. Question
1 point(s)
$(n+3)^2 > 2^{n+3} $ is true for ________________.
- all $ n \geq 3 $
- all $ n \geq 2 $
- all $ n $
- no $n \in N $
Correct

Incorrect

Hint

$P(n): (n+3)^2 > 2^{n+3}$
For $n=0$
$P(0): 3^2>2^3$
$: 9 > 8$ , it is true.
$P(1): 4^2>2^4$
$:16 > 16$ , it is false
$P(2): 5^2 > 2^5$
$:125 >32$ , it is false
$\therefore$ $P(n): (n+3)^3 > 2^{n+3}$ is false for $n=1,2,3,4,...$ i.e., $n\in N$ .
$\therefore$ no $n\in N$ i.e., $P(0)$ satisfy the above inequation.
Question 22 of 25

22. Question
1 point(s)
$3^{2n}-1 $ is divisible by ____________.
- 32
- 16
- 8
- 9
Correct

Incorrect

Hint

$f(n)= 3^{2n}-1$
For $n=1,f(1)= 3^2-1=9-1=8$
For $n=2,f(2)= 3^4-1=81-1=80$ , is divisible by $8$ .
$\therefore$ for $n=k,f(k)=3^{2k}-1$ is true.
For $n=k+1,$
$\begin{aligned} f(k+1)&=3^{2(k+1)}-1\\&=3^{2(k+1)}-1\\&=3^{2k+2}-1\\&=3^{2k}\cdot 3^2-1\\&=3^{2k}\cdot3^2-3^2+3^2-1\\&=3^2(3^{2k}-1)+9-1\\&=9(8m)+8\\&=8(9m+1)\end{aligned}$ , is divisible by $8$ .
$\therefore$ $f(k+1)$ is true.
$\therefore$ $f(n)=3^{2n}-1$ is divisible by $8$ .
Question 23 of 25

23. Question
1 point(s)
$a^{2n-1}+b^{2n-1} $ is divisible by ____________.
- $ a+b $
- $ (a+b)^2 $
- $ (a+b)^3 $
- $ a-b $
Correct

Incorrect

Hint

$f(n)= a^{2n-1}+b^{2n-1}$
For $n=1,f(1)= a^{2-1}+b^{2-1}=a+b$
For $n=2,f(2)= a^{4-1}+b^{4-1}=a^3+b^3=(a+b)(a^2-ab+b^2)$
$\therfore$ For $n=k, f(k)=a^{2k-1}+b^{2k-1}$ is true
For $n=k+1,f(k+1)= a^{2(k+1)-1}+b^{2(k+1)-1}$
$\begin{aligned} &=a^{2k+2-1}+b^{2k+2-1}=a^{2k-1} \cdot a^2 + b^{2k-1} \cdot b^2 \\ &= a^{2k-1} \cdot a^2+ a^2 b^{2k-1}-a^{2}b^{2k-1}+b^{2k-1} \cdot b^2\\&=a^2(a^{2k-1}+b^{2k-1})-b^{2k-1}(a^2-b^2) \\ &=a^2[(a+b)m]-b^{2k-1}(a-b)(a+b)\\&=(a+b)[a^2m-(a-b)\cdot b^{2k-1}]\end{aligned}$
$\therefore f(k+1)$ is true
$\therefore f(m)$ is true
Question 24 of 25

24. Question
1 point(s)
$P(n): 2^n(n-1)! $ < $n^n$ is true if
- n<2
- $n>2$
- $n \geq 2$
- never true
Correct

Incorrect

Hint

$P(n): 2^n (n-1) ! < n^n$
$P(1): 2<1$ is false
$P(2): 4<4$ is false
$\therefore 2^n (n-1) ! < n^n$ is true for $n>2$
Question 25 of 25

25. Question
1 point(s)
$3^{3n}-2n+1 $ is divisible by ____________.
- 2
- 3
- 4
- 5
Correct

Incorrect

Hint

$f(n)=3^{3n}-2n+1$
$f(1)=27-2+1=26=2 \times 13$
$f(2)=3^{6}-2 \times 2 +1=726=2 \times 343$
$f(k)$ is true for $n=k$
$f(k+1)$ is true for by induction hypothesis
$\therefore 3^{3n}-2n+1$ is divisible by $2$ .